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Jharkhand Board JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.2 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.2

Unless stated otherwise, use π = \(\frac{22}{7}\)

Question 1.
Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.
Solution :
Area of a sector = \(\frac{\pi r^2 \theta}{360}\)
r = 6, θ = 60°
= \(\frac{22}{7} \times \frac{6 \times 6 \times 60}{360}=\frac{132}{7}\) = 18.85 cm².

Question 2.
Find the area of a quadrant of a circle whose circumference is 22 cm.
Solution :

С = 2πr = 22

Question 3.
The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
Solution :
Length of the minute hand = 14 cm = r.
In 15 minutes the minute hand sweeps an area equal to a quadrant.
Area of the quadrant = \(\frac{\pi r^2}{4}\)

Alternative Method:
One minute = 6°
5 minutes = 30°
Area swept by the minute hand
= πr² \(\frac{θ}{360°}\)
= \(\frac{22}{7}\) × 14 × 14 × \(\frac{30°}{360°}\)
= 51.33 cm².

Question 4.
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding: (i) minor segment, (ii) major sector. (Use π = 3.14)
Solution :

Radius of the circle = 10 cm
Major segment is making 360° – 90° = 270°
Area of the sector making angle 270° = \(\frac{270°}{360°}\) × πr² cm²
= \(\frac{1}{4}\) × 10²π = 25 π cm²
= 25 × 3.14 cm² = 235.5 cm²
∴ Area of the major segment = 235.5 cm²
Height of ΔAOB = OA = 10 cm
Base of ΔAOB = OB = 10 cm
Area of ΔAOB = \(\frac{1}{2}\) × OA × OB
= \(\frac{1}{2}\) × 10 × 10 = 50 cm²
Major segment is making 90°
Area of the sector making angle 90° = \(\frac{90°}{360°}\) × πr² cm²
= \(\frac{1}{4}\) × 10²
= 25 × 3.14 cm² = 78.5 cm²
Area of the minor segment = Area of the sector making angle 90° – Area of ΔAOB
= 78.5 cm² – 50 cm² = 28.5 cm².

Question 5.
In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find: (i) the length of the arc, (ii) area of the sector formed by the arc, (iii) area of the segment formed by the corresponding chord.
Solution :

(i) Length of the arc AB = \(\frac{2 \pi r \theta}{360^{\circ}}\) θ = 60°, r = 21
= 2 × \(\frac{22}{7} \times \frac{21 \times 60}{360}\) = 22cm

(ii) Area of the sector formed by the arc = \(\frac{\pi r^2 \theta}{360^{\circ}}\)
= \(\frac{22}{7} \times \frac{21 \times 21 \times 60}{360^{\circ}}\)
= 11 × 21 = 231 cm².

(iii) Area of the segment formed APBLA = Area of the sector PAOB – Area of the ΔOAB
Area of ΔOAB = ?
From O, draw OL ⊥ AB. AL = ?, OL = ?

Question 6.
A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use π = 3.14 and \(\sqrt{3}\) = 1.73).
Solution :

Area of the minor segment APB = Area of the sector OAPB – Area of ΔOAB
= \(\frac{\pi r^2 \theta}{360^{\circ}}\) – Area of ΔOAB
Area of the sector OAPB = \(\frac{3.14 \times 15 \times 15 \times 60}{360}\) = 1.57 × 75 cm²
OPB is an isosceles triangle. OA = OB ∴ \(\hat{A}\) = \(\hat{B}\) = 60°.
The triangle becomes an equilateral triangle.
Area of the sector OAPB = 1.57 × 75 = 117.75 cm²
∴ Area of the minor segment APB = 117.75 – 97.31 = 20.44 sq.cm.
Area of the major segment AQBA = Area of the circle – Area of the minor segment
= πr² – 20.44
= (3.14 × 15 × 15) – 20.44
= 706.50 – 20.44 = 686.06 sq.cm.

Question 7.
A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use π = 3.14 and \(\sqrt{3}\) = 1.73).
Solution :

Area of the segment APB = Area of the sector PAOB – Area of ΔOAB
r = 12, θ = 120°, π = 3.14
∴ Area of the sector PAOB = \(\frac{\pi r^2 \theta}{360^{\circ}}\)
= \(\frac{3.14 \times 12 \times 12 \times 120}{360}\)
= 3.14 × 12 × 4
= 3.14 × 48
= 150.72 cm².
Area of ΔOAB = \(\frac{1}{2}\) × b × h b = AB, h = OC
Draw OC ⊥ AB.
AOB is an isosceles triangle. OC ⊥ AB.
ΔAOC ≅ BOC (RHS)
∴ AC = CB.
A\(\hat{O}\)C = 60°, O\(\hat{A}\)C = 30°, A\(\hat{C}\)O = 90°
In ΔOAC, O\(\hat{C}\)A = 90°. sin O\(\hat{A}\)C = \(\frac{OC}{OA}\)
sin 30° = \(\frac{OC}{12}\) sin 30° = \(\frac{1}{2}\)
\(\frac{1}{2}\) = \(\frac{OC}{12}\)
2OC = 12
OC = \(\frac{12}{2}\) = 6 cm.

In ΔOAC, O\(\hat{C}\)A = 90°
OC² + AC² = OA²
6² + AC² = 12²
AC² = 12² – 6²
= (12 + 6) (12 – 6)
= 18 × 6 = 108
AC = \(\sqrt{108}\) = \(\sqrt{36 \times 3}\) = 6\(\sqrt{3}\)
AC = CB = 6\(\sqrt{3}\)
∴ AB = 12\(\sqrt{3}\)
Area of ΔOAB = \(\frac{1}{2}\) × b × h
= \(\frac{1}{2}\) × AB × OC
= \(\frac{1}{2}\) × 12\(\sqrt{3}\) × 6 = 36\(\sqrt{3}\)
Area of the segment APB = Area of the sector PAOB – Area of triangle OAB
= 150.72 – 62.28
= 88.44 cm².

Question 8.
A horse is tied to a peg at one corner of a square-shaped grass field of side 15 m by means of a 5 m long rope. Find

(i) the area of that part of the field in which the horse can graze.
(ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use π = 3.14).
Solution :
(i) When the rope is 5m long, area grazed = \(\frac{\pi r^2}{4}\) (quadrant)
= \(\frac{3.14 \times 5 \times 5}{4}=\frac{78.5}{4}\) = 19.625 m².

(ii) When the rope is 10 m long, area grazed = \(\frac{\pi r^2}{4}\)
= \(\frac{3.14 \times 10 \times 10}{4}=\frac{3.14 \times 100}{4}=\frac{314}{4}\)
= 78.5 cm².
Increased area available when the rope is 10 m long is
= 78.500 – 19.625 = 58.875 cm².

Question 9.
A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in the figure. Find:
(i) the total length of the silver wire required
(ii) the area of each sector of the brooch.
Solution :

A\(\hat{O}\)B = 180°
It is divided into five equal parts,
∴ Each part = θ = \(\frac{180°}{5}\) = 36°.
Total length of silver wire used = πd × 5 × 35
= \(\frac{22}{7}\) × 35 + 175
= 110 + 175
= 285 mm².

Area of each sector = \(\frac{\pi r^2 \theta}{360^{\circ}}\)
= \(\frac{22}{7} \times \frac{35}{2} \times \frac{35}{2} \times \frac{36}{360}\)
= \(\frac{11 \times 35}{4}=\frac{385}{4}\) mm².

Question 10.
An umbrella has 8 ribs which are equally spaced. Assuming umbrella to be a flat circle of radius 45 cm, find the area between two consecutive ribs of the umbrella.
Solution :
Number of ribs in umbrella = 8
Radius of umbrella while flat = 45 cm.
Angle between two consecutive ribs of the umbrella = \(\frac{360}{8}\) = 45°

Question 11.
A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.
Solution :
Given, length of wiper blade = 25 cm = r (say)
Angle made by the blade, θ = 115°

Question 12.
To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use π = 3.14)
Solution :
π = 3.14, r = 16.5, θ = 80°
Area warned = \(\frac{\theta \pi r^2}{360^{\circ}}\)
= \(\frac{3.14 \times 16.5 \times 16.5 \times 80}{360}\)
= 3.14 × 5.5 × 11
= 3.14 × 60.5 = 189.970
= 189.97 sq.kms.

Question 13.
A round table cover has six equal designs as shown in the figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Re. 0.35 per cm². (Use \(\sqrt{3}\) = 1.7).

Solution :
[Draw a circle with centre O using a convenient radius. Draw a diameter AOD. Keep the protractor on AD and make three equal angles of 60° each such that A\(\hat{O}\)B = B\(\hat{O}\)C = C\(\hat{O}\)D. Produce BO and CO to meet the circumference at E and F respectively. Join AB, BC, CD, DE, EF, EA. We get a regular hexagon and six segments which are shaded.

Let one segment be APB.
Find the area of one segment. Multiply it by 6. We get the area of the six segments formed. Find the cost of making these designs as follows:
Area of one design × 6 × Rate]

The design APB is nothing but a segment. We have to find the area of this segment using the following relation:
Area of the sector OAPB – Area of equilateral ΔAOB

Cost of making these six designs = Area × Rate
= 464.8 × 0.35
= Rs. 162.68.

Question 14.
Tick the correct answer in the following:
Area of a sector of angle p (in degrees) of a circle with radius R is
(A) \(\frac{P}{180}\) × 2πR²
(B) \(\frac{P}{180}\) × πR²
(C) \(\frac{P}{360}\) × 2πR
(D) \(\frac{P}{720}\) × 2πR²
Solution :
Area of sector = \(\frac{\theta \pi r^2}{360^{\circ}}\) (Given, θ = p, r = R)
= \(\frac{p \pi R^2}{360}=\frac{p 2 \pi R^2}{720}\)

Jharkhand Board JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.1 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.1

Unless stated otherwise, use π = \(\frac{22}{7}\)

Question 1.
The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.
Solution :
Circumference of circle 1 = 2πr
= 2 × π × 19 cm = 2π × 19
Circumference of circle 2 = 2 × π × 9 cm = 2π × 9
Sum of the circumferences = (2π × 19) + (2π × 9)
= 2π(19 + 9)
= 2π 28
= 56 π.
Circumference of circle 3 = 56π.
2πr = 56π
r = \(\frac{56π}{2π}\)
= 28 cm.

Alternative Method:
r₁ = 19 cm, r₂ = 9 cm, R = ?
2πr₁ + 2πr₂ = 2πR (given)
2π(г₁ + r₂) = 2πR
(19 + 9) = R
∴ R = 28 cm.

Question 2.
The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.
Solution :
Area of circle 1 = π × 8² cm²
Area of circle 2 = π × r² = π × 6²
Sum of the areas = π(8² + 6²)
= π(64 + 36)
= 100π
Area of circle 3 = 100 π
πr² = 100 π
r² = \(\frac{100π}{π}\) = 100
∴ r = 10 cm.

Alternative Method:
πr₁² + πr₂² = πR²
π(r₁² + r₂²) = πR²
8² + 6² = R²
64 + 36 = R²
\(\sqrt{100}\) = R
∴ R = 10 cm.

Question 3.
The figure depicts an archery target marked with its five scoring regions from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.
Solution :

KG = GE = EC = CA = AO = OB = 10.5 cm
Area of the gold band = πr²
= \(\frac{22}{7} \times \frac{21}{2} \times \frac{21}{2}\)
r = \(\frac{AB}{2 }=\frac{21}{2}\) = 10.5
= \(\frac{33 \times 21}{2}=\frac{693}{2}\)
= 346.5 cm².

Area of the red band d = 21 + 10.5 + 10.5 = 42 cm.
= (\(\frac{22}{7} \times \frac{42}{2} \times \frac{42}{2}\)) – 346.5(area of the gold circle)
= (66 × 21) – 346.5
= 1386 – 346.5 = 1039.5 cm².

Area of the blue band d = 6 × 10.5 = 63
= (\(\frac{22}{7} \times \frac{63}{2} \times \frac{63}{2}\)) – area of the red circle
= \(\frac{99 \times 63}{2}\) – (66 × 21)
= \(\frac{6237}{2}\) – 1386
= 3118.5 – 1386.0 = 1732.5 cm²

Area of the black band = d = GH = 8 × 10.5 = 84
= (\(\frac{22}{7} \times \frac{84}{2} \times \frac{84}{2}\)) – area of the blue circle
= \(\frac{22}{7} \times \frac{84}{2} \times \frac{84}{2}\) – (\(\frac{22}{7} \times \frac{63}{2} \times \frac{63}{2}\))
= (132 × 42) – \(\frac{99 \times 63}{2}\)
= 5544.0 – 3118.5 = 2425.5 cm²

Area of the white band = d = KL = 10 × 10.5 = 105
= (\(\frac{22}{7} \times \frac{105}{2} \times \frac{105}{2}\)) – (132 × 42) (area of black circle)
= \(\frac{165 \times 105}{2}\) – (132 × 42)
= \(\frac{17325}{2}\) – 5544
= 8662.5 – 5544.0 = 3118.5 cm².

Question 4.
The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?
Solution :
Distance travelled in one revolution = Circumference of the wheel
= 2πr d = 80
= 2 × \(\frac{22}{7} \times \frac{80}{2}=\frac{22 \times 80}{7}\) cm
Distance travelled by the car in 10 min — ?
Speed of the car = 66 km/hr.
Speed of the car = \(\frac{66}{60}\) × 10 = 11 km/min.
= 11 × 1000 mts./min.

Question 5.
Tick the correct answer in the following and justify your choice: If the perimeter and the area of a circle are numerically equal, then the radius of the circle is
(A) 2 units
(B) π units
(C) 4 units
(D) 7 units.
Solution :
2πr = πr²
2r = r²
Solution :
(A). r = 2 units.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 11 Constructions Ex 11.2 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 11 Constructions Exercise 11.2

In each of the following, give also the justification of the construction:

Question 1.
Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.
Solution :

Steps of construction:
1. Draw a line segment OP = 10 cm.
2. With O as centre and 6 cm as radius draw a circle C₁.
3. Draw the perpendicular bisector of OP. Let x be the midpoint of OP.
4. With x as centre and xO or XP as radius draw circle C₂ to cut circle C₁ at A and B.
5. Join PA and PB.
PA and PB are the tangents to the circle from P.
P\(\hat{A}\)O = 90° (Angle in a semicircle)
∴ PA ⊥ radius OA.
∴ PA is a tangent to the circle.

Question 2.
Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.
Solution :

Steps of construction:
1. With O as centre and 4 cm as radius draw circle C₁.
2. With O as centre and 6 cm as radius draw circle C₂.
3. Take a point P on circle C₂.
4. Draw the perpendicular bisector of OP. Let x be the midpoint of OP.
5. With x as centre and xO or xP as radius draw circle C₃ to cut circle C₁ at A and B.
Join PA. PA is the tangent to circle C₁ from P.
O\(\hat{A}\)P = 90° (Angle in the semicircle)
∴ Radius OA ⊥ AP.
Hence PA is a tangent ∴ PA = 4.5 cm.
Verification by calculation:
In ΔOAP, O\(\hat{A}\)P = 90°. OA = 4 cm, OP = 6 cm.
OA² + AP² = OP²
4² + AP² = 6²
AP² = 6² – 4²
= (6 + 4) (6 – 4)
= 10 × 2 = 20
AP = \(\sqrt{20}\)
= \(\sqrt{4 \times 5}\) = 2\(\sqrt{5}\)
= 2 × 2.3 = 4.6 cm.

Question 3.
Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameters each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.
Solution :

Steps of construction:
1. Draw circle C₁ with centre O and radius = 3 cm.
2. Take a diameter LM in the circle.
3. Take a point P to the left of O at a distance of 7 cm.
4. Take a point Q to the right of O at a distance of 7 cm.
5. From P draw a tangent PS to C₁.
6. From Q draw a tangent RQ to C₂.
PS and RQ are tangents drawn to the circle from points P and Q respectively.

Question 4.
Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°.
Solution :

B\(\hat{A}\)O = 60°
∴ B\(\hat{A}\)C = 180° – 60° = 120°
Steps of construction:
1. Draw a circle of radius 5 cm with centre O.
2. At O make an angle BC = 120°.
3. At B and C draw perpendiculars to the radii OB and OC.
4. Let the perpendiculars meet at A.
AB and AC are tangents drawn to the circle from A such that the angle between them is 60°.

Question 5.
Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.
Solution :
AC and AD are tangents, drawn from centre A to C₂.
BE and BF are tangents drawn from centre B to C₁.
AC = AD = 7.2 cm, BE = BF = 6.8 cm.
Steps of construction:
1. Draw a line segment AB – 8 cm.
2. With A as centre and radius 4 cm draw circle C₁.
3. With B as centre and radius 3 cm draw circle C₂.

4. Draw the perpendicular bisector of AB. Let x be the midpoint of AB.
5. With x as centre and radius xA or xB draw circle C₃.
6. Let it cut C, at E and F and cut C₂ at C and D.
7. Join AC, AD, BE, BF.
AC and AD are tangents drawn from A, the centre of circle C₁ to circle C₂.
BE and BF are tangents drawn from B, the centre of circle C₂ to circle C₁.

Question 6.
Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.
Solution :

Steps of construction:
1. Construct ΔABC given AB = 6 cm, BC = 8 cm, A\(\hat{B}\)C = 90°.
2. From B draw perpendicular BD to AC.
3. Let O be the midpoint of BC.
4. With O as centre and OB or OC as radius, draw C₁ to pass through B, D and C.
5. Join AO.
6. Draw its perpendicular bisector. Let x be the midpoint of AO.
7. With x as centre and xA or xO as radius, draw circle C₂. Let it cut C₁ at E and B.
8. Join AE. AB is already joined.
AB and AE are tangents to circle C₁ from A.
AE = AB = 6 cm.

Question 7.
Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.
Solution :

A circle is drawn keeping the bangle on the paper. We have to find its centre.
Draw any two chords AB and CD in it.
Draw their perpendicular bisectors. The point of intersection O of these bisectors gives the centre of the circle.
(The perpendicular bisector of a chord passes through the centre).
Let P be the external point. Join PO. Draw the perpendicular bisector of PO. Let x be the midpoint.
With x as centre and xO or xP as radius draw circle C₂. Let it cut circle C₁ at Q and R.
Join PQ and PR. These are the tangents to the circle C₁.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 11 Constructions Ex 11.1 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 11 Constructions Exercise 11.1

In each of the following, give the justification of the construction also:

Question 1.
Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.
Solution :

Steps of construction:
1. Draw the line segment AB = 7.6 cm.
2. At A, below AB, make an angle BAx = 30°.
3. At B, above AB, make an angle ABy = 30°.
B\(\hat{A}\)x = A\(\hat{B}\)y = 30°
These are alternate angles. ∴ Ax || By.
4. With a convenient radius cut off five equal parts
AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ in Ax.
5. With the same radius cut off eight equal parts.
BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅ = B₅B₆ = B₆B₇ = B₇B₈.
6. Join A₅B₈. Let it cut AB at C.
AC : CB = 5 : 8.

In ΔACA₅ and CBB₈
1. A\(\hat{C}\)A₅ = B\(\hat{C}\)B₈ (V.O.A.)
2. C\(\hat{A}\)A₅ = A\(\hat{B}\)B₈ (Alternate angles)
3. C\(\hat{A}\)₅A = B\(\hat{B}\)₈C (Alternate angles)
Δs are equiangular.
∴\(\frac{\mathrm{AA}_5}{\mathrm{BB}_8}=\frac{\mathrm{CA}_5}{\mathrm{AB}_8}=\frac{\mathrm{CA}}{\mathrm{BC}}\)
∴ \(\frac{\mathrm{CA}}{\mathrm{BC}}=\frac{\mathrm{AA}_5}{\mathrm{BB}_8}=\frac{5}{8}\)

Question 2.
Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are \(\frac{2}{3}\) of the corresponding sides of the first triangle.
Solution :

Steps of construction:

1. Construct ΔABC given AB = 4 cm, BC = 5 cm, AC = cm.
2. At B, make an acute angle CBx.
3. Divide Bx into three equal parts with a convenient radius.
4. Join B₃C.
5. From B₂ draw a parallel to B₃C.
6. Let it cut BC at C’.
7. At C’ make angle A’C’B = ACB.

Join A’C’.
∴ ΔABC ||| A’BC’.

In ΔABC and ΔA’BC’,
1.A\(\hat{B}\)C = A’\(\hat{B}\)C’ (Common angle)
2. A\(\hat{C}\)B = A’\(\hat{C’}\)B (Corresponding angles)
3. B\(\hat{A}\)C = B\(\hat{A’}\)C’ (Remaining angles)
Δs are equiangular.
∴ \(\frac{\mathrm{AC}}{\mathrm{A}^{\prime} \mathrm{C}^{\prime}}=\frac{\mathrm{AB}}{\mathrm{A}^{\prime} \mathrm{B}^{\prime}}=\frac{\mathrm{BC}}{\mathrm{BC} \mathrm{C}^{\prime}}\) = \(\frac{3}{2}\) ∴ [latex]\frac{BC’}{BC}=\frac{2}{3}[/latex]

Question 3.
Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are \(\frac{7}{5}\) of the corresponding sides of the first triangle.
Solution :
Steps of construction:

1. Construct a triangle ABC given BC= 7 cm, AB = 5 cm, AC = 6 cm.
2. At B, below BC, make an acute angle CBx.
3. With a convenient radius cut off seven equal parts BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅ = B₅B₆ = B₆B₇.
4. Join B₅C.
5. From B₁ draw a parallel to B₅C to cut BC produced at C’.
6. At C’ draw a parallel to CA to meet BA produced at A’.

A’BC’ is the required triangle.
In ΔABC and ΔA’BC’

Question 4.
Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1\(\frac{1}{2}\) times the corresponding sides of the isosceles triangle.
Solution :

Steps of construction:

1. Draw a line segment BC = 8 cm.
2. Draw its perpendicular bisector.
3. Cut off DA = 4 cm. (altitude given)
4. Join AB and AC. ABC is the required triangle.
5. At B, below BC, draw an acute angle CBx.
6. With a convenient radius cut off three equal parts BB₁ = B₁B₂ = B₂B₃.
7. Join B₂C.
8. At B₃ draw a parallel to B₂C to meet BC extended at C’.
9. At C’ draw a parallel to CA to meet BA produced at A’.
10. Join A’C’. A’BC’ is the required triangle similar to ΔABC.

In ΔA’BC’ and ABC
1. A’\(\hat{B}\)C’ = A\(\hat{B}\)C (Common angle)
2. B\(\hat{A}\)‘C’ = B\(\hat{A}\)C (Corresponding angles)
3. A’\(\hat{C}\)‘B = A\(\hat{C}\)B
Δs are equiangular.
∴ \(\frac{\mathrm{A}^{\prime} \mathrm{C}^{\prime}}{\mathrm{AC}}=\frac{\mathrm{BC}}{\mathrm{BC}}=\frac{\mathrm{A}^{\prime} \mathrm{B}}{\mathrm{AB}}\)
\(\frac{BC’}{BC}\) = \(\frac{3}{2}\)
BC = 8 cm, BC’ = 8 × \(\frac{3}{2}\) = 12 cm.
BA = CA = 5.6 cm ∴ A’B = A’C’ = 5.6 × \(\frac{3}{2}\) = 8.4 cm.

Question 5.
Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are \(\frac{3}{4}\) of the corresponding sides of the triangle ABC.
Solution :

Steps of construction:

1. Construct ΔABC given BC = 6 cm, AB = 5 cm, A\(\hat{B}\)C = 60°.
2. At B, below BC, make an acute angle CBx.
3. In Bx cut off four equal parts BB₁ = B₁B₂ = B₂B₃ = B₃B₄
4. Join B₄C.
5. From B₃ draw a parallel to B₄C to meet BC at C’.
6. At C’ draw a parallel to CA to meet CA at A’.

A’BC’ is the required triangle similar to ΔABC.
In Δs A’BC’ and ABC

Question 6.
Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are \(\frac{4}{3}\) times the corresponding sides of ΔABC.
Solution :

\(\hat{A}\) + \(\hat{B}\) + \(\hat{C}\) = 180°
\(\hat{A}\) + \(\hat{B}\) = 150° (105° + 45°)
∴ \(\hat{C}\) = 30°
Steps of construction:
1. In ΔABC, BC = 7 cm.
\(\hat{B}\) = 45°, \(\hat{C}\) = 105°
∴ \(\hat{A}\) = 180° – 45° – 105°
= 180° – 150° = 30°.
Construct ΔABC given BC = 7 cm, \(\hat{B}\) = 45°, \(\hat{C}\) = 30°.
2. At B draw an acute angle CBx.
3. In Bx cut off four equal parts BB₁ = B₁B₂ = B₂B₃ = B₃B₄ with a convenient radius.
4. Join B₃C.
5. At B₄ draw a parallel to B₃C to meet BC produced at C’.
6. At C’ make angle of 30° equal to A\(\hat{C}\)B. Let it meet BA produced at A’.
A’BC’ is the required triangle similar to ΔABC.
In ΔC’BB₄ CB₃ || C’B₄

Question 7.
Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are \(\frac{5}{3}\) times the corresponding sides of the given triangle.
Solution :

BC’ = \(\frac{5}{3}\) × BC = \(\frac{5}{3}\) × 3 = 5 cm
BA’= \(\frac{5}{3}\) × 4 = \(\frac{20}{3}\) = 6.6 cm
A’C’ = \(\frac{5}{3}\) × 5 = \(\frac{25}{3}\) = 8.3 cm.

Steps of construction:

1. Construct ΔABC given BC= 3 cm, \(\hat{B}\) = 90°, BA = 4 cm.
2. At B, make an acute angle CBx.
3. Cut off five equal parts BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅ along Bx with a convenient radius.
4. Join B₃C.
5. At B₅ draw a parallel to B₃C to meet BC produced at C’.
6. At C’ draw a parallel to CA to meet BA produced at A’.

A’BC’ is the required triangle similar to ΔABC.
BC’ : BC = BB₅ = BB₃
\(\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{BB}_5}{\mathrm{BB}_3}\) = \(\frac{5}{3}\)

In Δs A’BC’ and ABC
1. A’\(\hat{B}\)C’ = A\(\hat{B}\)C = 90°
2. B\(\hat{A}\)‘C’ = B\(\hat{A}\)C (Corresponding angles)
3. B\(\hat{C}\)‘A’ = B\(\hat{C}\)A
Δs are equiangular.
∴ \(\frac{A’C’}{AC}\) = \(\frac{BC’}{BC}\) = \(\frac{BA’}{BA}\) = \(\frac{5}{3}\)

Jharkhand Board JAC Class 10 Maths Solutions Chapter 10 Circles Ex 10.2 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 10 Circles Exercise 10.2

Question 1.
From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
(A) 7 cm
(B) 12 cm
(C) 15 cm
(D) 24.5 cm
Solution :

OQ² = TO² – TQ²
= 25² – 24²
= (25 + 24) (25 – 24)
= 49 × 1 = 49
OQ = \(\sqrt{49}\) = 7 cm.

Question 2.
In the figure, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to
(A) 60°
(B) 70°
(C) 80°
(D) 90°
Solution :

POQT is a cyclic quadrilateral because \(\hat{\mathbf{P}}\) + \(\hat{\mathbf{O}}\) = 180°.
∴ \(\hat{\mathbf{O}}\) + \(\hat{\mathbf{T}}\) = 180°
latex]\hat{\mathbf{O}}[/latex] = 110°.
latex]\hat{\mathbf{T}}[/latex] = 180° – 110° = 70°.

Question 3.
If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80°, then ∠POA is equal to

A) 50°
B) 60°
C) 70°
D) 80°
Solution :
In ΔPAO, P\(\hat{A}\)O = 90° A\(\hat{P}\)O = 40°
∴ A\(\hat{O}\)P = 180° – 90° – 40° = 50°.

Question 4.
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Solution :

Data: O is the centre of the circle.
AB is a diameter. CD and EF are tangents drawn at A and B.
To prove: CD || EF.
Proof: O\(\hat{A}\)C = 90°
O\(\hat{B}\)E = 90°
B\(\hat{A}\)C + A\(\hat{B}\)E = 180°
These are co-interior angles.
Radius is ⊥ to the tangent at the point of contact

Question 5.
Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Solution :

Let us consider a circle with centre O.
Let AB be a tangent which touches the circle at P.
We have to prove that the line perpendicular to AB at P passes through the centre O.
We shall prove this by contradiction method.
Let us assume that the perpendicular to AB at P does not pass through centre O.
Let it pass through another point O’. Join OP and O’P.
As the perpendicular to AB at P passes through O’, therefore,
∠O’PB = 90° ……..(1)
O is the centre of the circle and P is the point of contact. We know that the line joining the centre and the point of contact to the tangent of the circle are perpendicular to each other.
∴ ∠OPB = 90° ……..(2)
Comparing equations (1) and (2), we obtain
∠O’PB = ∠OPB ….(3)
From the figure, it can be observed that,
∠O’PB < ∠OPB ………..(4)
Therefore, ∠O’PB = ∠OPB is not possible. It is only possible when the line O’P coincides with OP. Therefore, the perpendicular to AB at P passes through the centre O.

Question 6.
The length of a tangent from a point A at a distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

Solution :
A\(\hat{B}\)O = 90° (The radius is perpendicular to the tangent at the point of contact)
BO² + AB² = AO²
BO² = AO² – AB²
= 5² – 4²
= 25 – 16 = 9
BO = \(\sqrt{9}\) = 3 cm.
Hence, radius of the circle is 3 cm.

Question 7.
Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Solution :

O is the centre of the concentric circles C₁ and C₂.
AB is a chord of the bigger circle C₁.
AB is a tangent to circle C₂.
OC = 3 cm, OB = 5 cm.
To find AB.
Proof: In ΔOCB,
O\(\hat{C}\)B = 90° (The tangent AB to C₂ is ⊥ to the radius CO drawn from the point of contact C)
In ΔOCB, O\(\hat{C}\)B = 90°
OC² + CB² = OB²
3² + CB² = 5²
CB² = 5² – 3²
= 25 – 9 = 16
CB = 4 cm.
In circle C₁ AB is a chord. OC ⊥ AB.
∴ AC = CB = 4 cm. The ⊥ drawn to a chord from the centre bisects the chord.
AB = AC + CB
= 4 + 4 = 8 cm.

Question 8.
A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC.
Solution :

A, B, C and D are external points. Tangents drawn to the circle from these points are equal.
AP = AS ….(1) (Tangents from A)
BP = BQ ….(2) (Tangents from B)
CR = CQ ….(3) (Tangents from C)
DR = DS ….(4) (Tangents from D)
Adding the results
AP + BP + CR + DR = AS + BQ+CQ + DS
AB + CD = AS + DS + BQ +CQ
∴ AB + CD = AD + BC.

Question 9.
In the figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B. Prove that ∠AOB = 90°.
Solution :

Given: XY || X’Y’.
XPA, X’QY’ and AB are the tangents.
To prove: ∠AOB = 90°
Construction: Join OC.
Proof: In ΔOAP and ΔOAC
AP = AC (Theorem 4.2)
OA = OA (Common)
OP = OC (Radii of circle)
AOAP AOAC (S.S.S. congruency)
Since XY || X’Y’
∠PAB + ∠QBA = 180° (Co-interior angles)
\(\frac{1}{2}\)∠PAB + \(\frac{1}{2}\)∠QBA = 90°
∴ ∠OAB + ∠OBA = 90°
In ΔAOB, ∠AOB = 180° – (∠OAB + ∠OBA)
= 180° – 90° = 90°
∴ ∠AOB = 90°.
The line joining the external point to the centre, bisects the angle between the tangents.

Question 10.
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
Solution :

Data: O is the centre of the circle.
AB and AC are tangents drawn from A.
To prove: B\(\hat{\mathbf{A}}\)C + B\(\hat{\mathbf{O}}\)C = 180°.
Proof: AB is a tangent. BO is the radius drawn from the point of contact B.
∴ O\(\hat{\mathbf{B}}\)A = 90°
AC is a tangent. CO is the radius drawn from the point of contact C.
∴ O\(\hat{\mathbf{C}}\)A = 90°
∴ O\(\hat{\mathbf{B}}\)A + O\(\hat{\mathbf{C}}\)A = 90°.
∴ B\(\hat{\mathbf{O}}\)C + B\(\hat{\mathbf{A}}\)C = 180° [Sum of angles of a quadrilateral is 360°]

Question 11.
Prove that the parallelogram circumscribing a circle is a rhombus.
Solution :

Data: ABCD is a circumscribed parallelogram.
AB = CD, AD = BC
To prove: ABCD is a rhombus.
AB = BC – CD – DA
Proof: ABCD is a circumscribed quadrilateral.
∴ AB + CD = AD + BC (Proved in problem 8)
AB = DC, AD = BC (Opposite sides of the parallelogram)
AB + AB = AD + AD
2AB = 2AD
AB = AD
AB = CD, AD = BC
AB = BC = CD = DA. [Since adjacent sides of a parallelogram are equal]
∴ ABCD is a rhombus.

Question 12.
A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC.

Solution :

In the adjoining figure, the circle touches AB at F, BC at D, CA at E.
AE = AF = x. (Tangents to the circle from A)
Tangent BF = Tangent BD = 8 cm
Tangent CE = Tangent CD = 6 cm
AB = AF + FB = (x + 8) cm
AC = AE + EC = (x + 6) cm
CB = CD + DB = 6 + 8 = 14 cm.
2S(perimeter of AABC) = AB + BC + CA
= (x + 8) + (6 + 8) + (x + 6).
= 2x + 28
S = x + 14

Area of ΔABC = \(\frac{1}{2}\) × base × altitude
= \(\frac{1}{2}\) × 14 × 4 = 28 cm².
Area of ΔOAB = \(\frac{1}{2}\) × 4(x + 8) = 2(x + 8)
Area of ΔOAC = \(\frac{1}{2}\) × 4(x + 6) = 2(x + 6)
Area of ΔOBC = \(\frac{1}{2}\) × 4 × 14 = 28 cm.
Total area = Area of ΔOBC + Area of ΔOAB + Area of ΔOAC
= 28 + 2x + 12 + 2x + 16
∴ 4x + 56 = 4(x + 14)
∴ 4\(\sqrt{3 x^2+42 x}\) = 4(x + 14)
\(\sqrt{3 x^2+42 x}\) = x + 14
Squaring both sides 3x² + 42x = (x + 14)² = x² + 28x + 196
= 2x² + 145 – 196 = 0 x² + 7x – 98 = 0
∴ (x + 14) (x – 7) = 0 x = – 14 or x = 7
AC = x + 8
= 7 + 8
= 15 cm.

AC = x + 6
= 7 + 6
= 13 cm.

BC = 6 + 8
= 14 cm.

Question 13.
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Solution :

Construction: Join OP, OQ, OR and OS.
Proof: The two tangents drawn from an external point to a circle subtend equal angles at the centre.
∴ ∠1 = ∠2, ∠3 = ∠4, ∠5 = ∠6 and ∠7 = ∠8.
Now, ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360° (Sum of angles subtended at a point)
2(∠2 + ∠3 + ∠6 + ∠7) = 360° and
2(∠1 + ∠8 + ∠4 + ∠5) = 360°
⇒ (∠2 + ∠3) + (∠6 + ∠7) = 180°
(∠1 + ∠8) + (∠4 + ∠5) = 180°
∴ ∠AOB + ∠COD = 180° and ∠AOD + ∠BOC = 180°.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 10 Circles Ex 10.1 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 10 Circles Exercise 10.1

Question 1.
How many tangents can a circle have?
Solution :
A circle can have infinite number of tangents.

Question 2.
Fill in the blanks:
(i) A tangent to a circle intersects it in ________ point (s).
(ii) A line intersecting a circle in two points is called a _______.
(iii) A circle can have _________ parallel tangents at the most.
(iv) The common point of a tangent to a circle and the circle is called ________.
Solution :
(i) one,
(ii) sccant,
(iii) two,
(iv) point of contact.

Question 3.
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is:

(A) 12 cm
(B) 13 cm
(C) 8.5 cm
(D) \(\sqrt{119}\) cm.
Solution :
PQ² = OQ² – OP² (Using Pythagoras theorem)
= (12)² – (5)²
= 144 – 25
= 119.
PQ = \(\sqrt{119}\) cm.

Question 4.
Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.
Solution :
AB is the given line. CD is the secant and PQ is the tangent to the circle at point R.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 9 Some Applications of Trigonometry Ex 9.1 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 9 Some Applications of Trigonometry Exercise 9.1

Question 1.
A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30°.
Solution :

sin \(\hat{\mathrm{C}}\) = \(\frac{AB}{AC}\)
sin 30° = \(\frac{1}{2}\)
\(\frac{1}{2}\) = \(\frac{AB}{20}\)
AB = \(\frac{20}{2}\) = 10 m

Question 2.
A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.
Solution :

Question 3.
A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?
Solution:

(i) Figure (a) shows the slide for children below the age of 5 years.
Let BC = 1.5 m be the height of the slide. Slide AC is inclined at CAB = 30° to the ground.
In right angled ΔABC, sin 30° = \(\frac{BC}{AC}\)
\(\frac{1}{2}=\frac{15}{AC}\)
⇒ AC = 3 m.

(ii) Figure (b) shows the slide for elder children. Let RQ = 3 m be the height of the slide. Slide PR is inclined at ∠RPQ = 60° to the ground.
In right angled ΔPQR, sin 60° = \(\frac{RQ}{PR}\) ⇒ \(\frac{\sqrt{3}}{2}\) = \(\frac{3}{PR}\)
PR = \(\frac{3 \times 2}{\sqrt{3}}\) = 2\(\sqrt{3}\) m.

Question 4.
The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.
Solution :

tan C = \(\frac{AB}{CB}\)
tan 30° = \(\frac{1}{\sqrt{3}}\)
tan 30° = \(\frac{h}{30}\)
\(\frac{1}{\sqrt{3}}\) = \(\frac{h}{30}\)
h\(\sqrt{3}\) = 30
h = \(\frac{30}{\sqrt{3}}=\frac{30 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}=\frac{30 \sqrt{3}}{3}\)
h = 10\(\sqrt{3}\)mts.
Hence, height of the tower is 10\(\sqrt{3}\) mts.

Question 5.
A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.
Solution :

Length of the string is 40\(\sqrt{3}\) mts.

Question 6.
A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.
Solution :

Let the boy be standing at point B initially. He walks towards the building and reaches point D. From the figure, the distance walked by the boy towards the building is BD.
AC = AG – CG
AC = 30 – 1.5 = 28.5
Now, in ΔABC, we have
tan 30 = \(\frac{AC}{BC}\)
BC = \(\frac{AC}{tan 30}\)
BC = 28.5\(\sqrt{3}\)
Again, in ΔADC, we have
tan 60 = \(\frac{AC}{DC}\)
DC = \(\frac{AC}{tan 60}\)
DC = \(\frac{28.5}{\sqrt{3}}\)
DC = \(\frac{28.5 \sqrt{3}}{3}\) = 9.5\(\sqrt{3}\)
BD = BC – DC
BD = 28.5\(\sqrt{3}\) – 9.5\(\sqrt{3}\)
BD = 19\(\sqrt{3}\)
The distance walked by the boy towards the building is 19\(\sqrt{3}\) m.

Question 7.
From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.
Solution :

tan 60° = \(\sqrt{3}\)
tan 45° = 1
In ΔADC, A\(\hat{\mathrm{D}}\)C = 60
tan 60° = \(\frac{AC}{AD}\)
\(\sqrt{3}\) = \(\frac{x+20}{DA}\)

In ΔBDA, A\(\hat{\mathrm{D}}\)B = 45°
tan 45° = \(\frac{AB}{AD}\) = 1
AB = AD = 20 mts.
DA\(\sqrt{3}\) = x + 20
\(\sqrt{3}\)DA = x + 20
\(\sqrt{3}\)(20) = x + 20
x = 20\(\sqrt{3}\) – 20
= 20(\(\sqrt{3}\) – 1)
Height of the tower B = 20(\(\sqrt{3}\) – 1) mts.

Question 8.
A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.
Solution :

tan 60° = \(\sqrt{3}\)
In ΔADC, tan 60° = \(\frac{AC}{CD}\)
\(\sqrt{3}\) = \(\frac{AC}{CD}\)
AC = \(\sqrt{3}\)CD

In ΔBDC, tan B\(\hat{\mathrm{D}}\)C = tan 45° = \(\frac{BC}{CD}\)
1 = \(\frac{BC}{CD}\)
CD = BC.

A = AC – CB
= AC – CD (∵ CB = CD)
= \(\sqrt{3}\)CD – CD
1.6 = CD(\(\sqrt{3}\) – 1)

Height of the pedestal = 0.8(\(\sqrt{3}\) + 1) mts.

Question 9.
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.
Solution :

Let the height of the building CD be h.
tan 60° = \(\sqrt{3}\)
tan 30° = \(\frac{1}{\sqrt{3}}\)
In ΔABC, tan \(\hat{\mathrm{C}}\) = \(\frac{AB}{AC}\)
tan 60° = \(\frac{50}{AC}\)
\(\sqrt{3}\) = \(\frac{50}{AC}\)
AC\(\sqrt{3}\) = 50°
AC = \(\frac{50}{\sqrt{3}}\)mts.

Height of the building = 16\(\frac{2}{3}\) mts.

Question 10.
Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.
Solution :

Let AE = x.
EB = 80 – x
tan 60° = \(\sqrt{3}\)
tan 30° = \(\frac{1}{\sqrt{3}}\)

Question 11.
A TV tower stands vertically on the bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the canal.
Solution :

tan 60° = \(\sqrt{3}\)
tan 30° = \(\frac{1}{\sqrt{3}}\)
In ΔABC, tan 60° = \(\frac{AB}{BC}\) = \(\frac{h}{x}\)
\(\frac{h}{x}\) = \(\sqrt{3}\)
∴ h = \(\sqrt{3}\)x
In ΔADB, tan 30° = \(\frac{AB}{DB}\)

Height = x\(\sqrt{3}\) = 10\(\sqrt{3}\) mts. Width of the canal = 10 mts.

Question 12.
From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.
Solution :

In ΔDBC, tan 60° = \(\frac{DC}{BC}\)
\(\sqrt{3}\) = \(\frac{DC}{BC}\)
= \(\frac{DC}{AE}\) = \(\frac{DC}{7}\) (BC = AE).

In ΔABE, tan 45° = \(\frac{AB}{AE}\)
1 = \(\frac{7}{AE}\)
∴ AE = 7 mts.
AB = BC = 7 mts.
DC = 7\(\sqrt{3}\)
∴ DE = DC + CE
= 7\(\sqrt{3}\) + 7 = 7(\(\sqrt{3}\) + 1) mts.

Question 13.
As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
Solution :

Hence, the distance between the two ships is 75(\(\sqrt{3}\) – 1) m.

Question 14.
A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30°. Find the distance travelled by the balloon during the interval.

Solution :

Let the initial position of the balloon be A and final position be B.
Height of the balloon above the girl’s height = 88.2 m – 1.2 m = 87m
Distance travelled by the balloon = DE = CE – CD

Distance travelled by the balloon, DE = CE – CD
= (87\(\sqrt{3}\) – 29\(\sqrt{3}\)) m
= 29\(\sqrt{3}\) (3 – 1) m
= 58\(\sqrt{3}\) m.

Question 15.
A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.
Solution :

In ΔBCA, tan 30° = \(\frac{h}{CA}\)
\(\frac{1}{\sqrt{3}}\) = \(\frac{h}{CA}\)
CA = h\(\sqrt{3}\) mts.

Question 16.
The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.
Solution :

Let AB be the tower. ∠ABC = x
∴ ∠ADB = 90° – x
In ΔABC tan x = \(\frac{AB}{BC}\)
tan x = \(\frac{AB}{4}\) ……………(i)
In ΔADB tan (90° – x) = \(\frac{AB}{9}\)
cot x = \(\frac{AB}{9}\) ……………(ii)
(i) × (ii)
tan x × cot x = \(\frac{AB}{4}\) × \(\frac{AB}{9}\)
tan x × \(\frac{1}{tan x}\) = \(\frac{\mathrm{AB}^2}{36}\)
1 = \(\frac{\mathrm{AB}^2}{36}\)
AB² = 36
AB = ± \(\sqrt{36}\)
AB = ± 6
∴ Height of the tower AB = 6 m.
Note: C and D can be taken on the same side of AB.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.4 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 8 Introduction to Trigonometry Exercise 8.4

Question 1.
Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Solution :
(i) We know that, cosec²A = 1 + cot² A ⇒ cosec A = \(\sqrt{1+\cot ^2 \mathrm{~A}}\)

Question 2.
Write all the other trigonometric ratios of ∠A in terms of sec A.
Solution :
(i) sin A sin² A + cos² A = 1
sin² A = 1 – cos² A

Question 3.
Evaluate:
(i) \(\frac{\sin ^2 63^{\circ}+\sin ^2 27^{\circ}}{\cos ^2 17^{\circ}+\cos ^2 73^{\circ}}\)
(ii) sin 25° cos 65° + cos 25° sin 65°.
Solution :
(i) \(\frac{\sin ^2 63^{\circ}+\sin ^2 27^{\circ}}{\cos ^2 17^{\circ}+\cos ^2 73^{\circ}}\)
sin (90 – θ) = cos θ
sin (90 – 27°) = cos 27°
sin 63° = cos 27°
sin² 63 = cos² 27°

cos (90 – θ) = sin θ
cos (90° – 73°) = sin 73°
cos (17°) = sin 73°
cos² 17° = sin² 73°

(ii) sin 25° cos 65° + cos 25° sin 65°
sin (90° – θ) = cos θ
sin (90° – 25°) = cos 25°
sin (90° – 65°) = cos 65°
sin 25° = cos 65°

cos (90° – θ) = sin θ
cos (90° – 65°) = sin 65°
cos 25° = sin 65°

sin 25° = cos 65°
cos 65° . cos 65° + sin 65° . sin 65°
∴ cos² 65 + sin² 65 = 1

Question 4.
Choose the correct option. Justify your choice.
(i) 9 sec² A – 9 tan² A=
(A) 1
(B) 9
(C) 8
(D) 0
Solution :
9 sec² A – 9 tan² A
9(1 + tan² A) – 9 tan² A
9 + 9 tan² A – 9 tan² A = 9.
∴ 9 sec² A – 9 tan² A = 9.

(ii) (1 + tan θ + sec θ) (1 + cot 0 – cosec θ) =
(A) 0
(B) 1
(C) 2
(D) – 1
Solution :

(iii) (sec A + tan A) (1 – sin A) =
(A) sec A
(B) sin A
(C) cosec A
(D) cos A
Solution :

(iv) \(\frac{1+\tan ^2 A}{1+\cot ^2 A}\) =
(A) sec² A
(B) – 1
(C) cot² A
(D) tan² A
Solution :

Question 5.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
Solution :

(viii) (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A.

LHS = sin² A + cosec² A + 2 sin A cosec A + cos² A + sec² A + 2 cos A sec A [∵ (a + b)² = a² + b² + 2ab]
= (sin² A + cos² A) + (1 + cot² A) + 2 sin A . \(\frac{1}{sin A}\) + (1 + tan² A) + 2 cos A . \(\frac{1}{cos A}\) (∵ 1 + cot² A = cosec² A and sec² A = 1 + tan² A)
= 1 + 1 + cot² A + 2 + 1 + tan² A + 2 (∵ sin² A + cos² A = 1)
= 7 + tan² A + cot² A = RHS.

(ix) (cosec A – sin A) (sec A – cos A) = \(\frac{1}{\tan A+\cot A}\)
Solution :
[Hint: Simplify LHS and RHS separately].
LHS = (cosec A – sin A) (sec A – cos A)

Jharkhand Board JAC Class 10 Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.3 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 8 Introduction to Trigonometry Exercise 8.3

Question 1.
Evaluate:
(i) \(\frac{\sin 18^{\circ}}{\cos 72^{\circ}}\)
(ii) \(\frac{\tan 26^{\circ}}{\cot 64^{\circ}}\)
(iii) cos 48° – sin 42°
(iv) cosec 31° – sec 59°.
Solution :
\(\frac{\sin 18^{\circ}}{\cos 72^{\circ}}\) sin θ = cos (90° – θ)
sin 18° = cos (90° – 18°) = cos 72°
∴ \(\frac{sin 18}{cos 72}\) = \(\frac{cos 72°}{cos 72°}\) = 1

(ii) \(\frac{\tan 26^{\circ}}{\cot 64^{\circ}}\) tan θ = cot (90 – θ)
tan 26 = cot (90 – 26) = cot 64°.
∴ \(\frac{tan 26°}{cot 64°}\) = \(\frac{cot 64}{cot 64}\) = 1

(iii) cos 48° – sin 42° = cos (90° – 42°) – sin 42°
= sin 42° – sin 42° = 0.

(iv) cosec 31° – sec 59° . cosec θ = sec (90 – θ)
cosec 31 = sec (90 – 31) = sec 59°.
cosec 31° – sec 59° = sec 59° – sec 59° = 0.

Question 2.
Show that:
(i) tan 48° tan 23° tan 42° tan 67° = 1.
(ii) cos 38° cos 52° – sin 38° sin 52° = 0.
Solution :
(i) tan 48° tan 23° tan 42° tan 67° = 1.
LHS = tan 48° tan 23° tan 42° tan 67°
= tan 48° . tan 23° tan (90° – 48°) . tan (90° – 23°)
= tan 48° . tan 23° . cot 48° . cot 23° [∵ tan (90° – θ) = cot θ]
= tan 48° . tan 23° . \(\frac{1}{tan 48°}\) . \(\frac{1}{tan 23°}\) [∵ cot θ = \(\frac{1}{tan θ}\)]
= 1 = RHS.

(ii) cos 38° cos 52° – sin 38° sin 52° = 0.
LHS = cos 38° . cos 52° – sin 38° sin 52°
= cos (90° – 52°) cos (90° – 38°) – sin 38°. sin 52°
= sin 52° . sin 38° – sin 38° . sin 52° [∵ cos (90° – θ) = sin θ]
= 0 = RHS.

Question 3.
If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.
Solution :
tan θ = cot (90° – θ)
tan 2A = cot (90° – 2A)
tan 2A = cot (A – 18°) ∵ cot (90° – 2A) = cot (A – 18°)
90° – 2A = A – 18
– 2A – A = – 18 – 90
– 3A = – 108
A = \(\frac{-108}{-3}\) = 36°.

Question 4.
If tan A = cot B, prove that A + B = 90°.
Solution :
tan A = cot B (90° – A)
tan A = cot B
∴ cot (90° – A) = cot B
∴ 90° – A = B
90° = B + A
A + B = 90°.

Question 5.
If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
Solution :
sec 4A = cosec (A – 20°)
cosec (90° – 4A) = cosec (A – 20°)
90° – 4A = A – 20°
– 4A – A = – 110°
– 5A = – 110°
A = 22°.

Question 6.
If A, B and C are interior angles of a triangle ABC, then show that
Solution :
sin (\(\frac{B+C}{2}\)) = cos \(\frac{A}{2}\)
sin θ = cos (90° – θ)

Question 7.
Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Solution :
sin 67° + cos 75°
= sin (90° – 23°) + cos (90° – 15°)
= cos 23° + sin 15°. (Since cos (90° – θ) = sin θ and sin (90° – θ) = cos θ)

Jharkhand Board JAC Class 10 Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.1 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 8 Introduction to Trigonometry Exercise 8.1

Question 1.
Evaluate the following:
(i) sin 60° cos 30° + sin 30° cos 60°
(ii) 2 tan² 45° + cos² 30° – sin² 60°

Solution :
(i) sin 60° cos 30° + sin 30° cos 60°
\(\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}\) + \(\frac{1}{2} \times \frac{1}{2}\)
\(\frac{3}{4}+\frac{1}{4}\) = 1

(ii) 2 tan² 45° + cos² 30° – sin² 60°
2(tan 45)² + (cos 30)² – (sin 60)²
2(1)² + (\(\frac{\sqrt{3}}{2}\))² – (\(\frac{\sqrt{3}}{2}\))² = 2

Question 2.
Choose the correct option and justify your choice:
(i) \(\frac{2 \tan 30^{\circ}}{1+\tan ^2 30^{\circ}}\)
(A) sin 60°
(B) cos 60°
(C) tan 60°
(D) sin 30°
Solution :

(ii) \(\frac{1-\tan ^2 45^{\circ}}{1+\tan ^2 45^{\circ}}\) =
(A) tan 90°
(B) 1
(C) sin 45°
(D) 0
Solution :
\(\frac{1-(-1)^2}{1+(1)^2}=\frac{0}{2}\) = 0

(iii) sin 2A = 2 sin A is true when A =
(A) 0°
(B) 30°
(C) 45°
(D) 60°
Solution :
sin 2A = sin 0° = 0
2 sin A = 2 sin 0° = 2(0) = 0.

(iv) \(\frac{2 \tan 30^{\circ}}{1-\tan ^2 30^{\circ}}\) =
(A) cos 60°
(B) sin 60°
(C) tan 60°
(D) sin 30°
Solution :

Question 3.
If tan (A + B) = \(\sqrt{3}\) and tan (A – B) = \(\frac{1}{\sqrt{3}}\), 0° < A + B ≤ 90°; A > B, find A and B.
Solution :
tan(A + B) = \(\sqrt{3}\)
tan 60° = \(\sqrt{3}\)
∴ A + B = 60°
tan (A – B) = \(\frac{1}{\sqrt{3}}\)
tan 30° = \(\frac{1}{\sqrt{3}}\)
∴ A – B = 30°
∴ \(\hat{\mathbf{A}}\) = 45°, \(\hat{\mathbf{B}}\) = 15°

A + B = 60°
45 + B = 60°
B = 60 – 45 = 15
∴ B = 15°

Question 4.
State whether the following are true or false. Justify your answer.
(i) sin (A + B) = sin A + sin B.
(ii) The value of sin θ increases as θ increases.
(iii) The value of cos θ increases as θ increases.
(iv) sin θ = cos θ for all values of θ.
(v) cot A is not defined for A = 0°.
Solution :
i) Let A = 30°, B = 60°.
sin 30° = \(\frac{1}{2}\)
sin 30 + sin 60°
sin (A + B) = sin 90° = 1
sin 60° = \(\frac{\sqrt{3}}{2}\)
\(\frac{1}{2}+\frac{\sqrt{3}}{2}\) = \(\frac{1+\sqrt{3}}{2}\)
sin (A+B) sin A + sin B
False.

iv) [θ = 0] sin 0 = 0
cos 0 = 1

θ = 30° sin 30° = \(\frac{1}{2}\)
cos 30° = \(\frac{\sqrt{3}}{2}\)

θ = 45° sin 45° = \(\frac{1}{\sqrt{2}}\)
cos 45° = \(\frac{1}{\sqrt{2}}\)

θ = 60° sin 60° = \(\frac{\sqrt{3}}{2}\)
cos 60° = \(\frac{1}{2}\)

θ = 90° sin 90° = 1
cos 90° = 0
False, because it is true only for θ = 45°.

(v) cot A = \(\frac{cos A}{sin A}\) . cot 0° = \(\frac{cos 0°}{sin 0°}\) = \(\frac{1}{0}\) = Undefined. True.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.1 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 8 Introduction to Trigonometry Exercise 8.1

Question 1.
In ΔABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:
(i) sin A, cos A, (ii) sin C, cos C.
Solution :

(i) sin A, cos A
AC² = AB² + BC²
= 24² + 7² = 576 + 49
= 625
∴ AC = 25

Question 2.
In the figure, find tan P – cot R.
Solution :

QR² = PR² – PQ² = 13² – 12²
= 169 – 144 = 25.
∴ QR = 5.
tan P = \(\frac{\mathrm{QR}}{\mathrm{PQ}}=\frac{5}{12}\)
cot R = \(\frac{\mathrm{QR}}{\mathrm{QP}}=\frac{5}{12}\)
∴ tan P – cot R = \(\frac{5}{12}=\frac{5}{12}\) = 0

Question 3.
If sin A = \(\frac{3}{4}\), calculate cos A and tan A.
Solution:

Given, sin A = \(\frac{3}{4}\) ⇒ \(\frac{P}{H}\) = \(\frac{3}{4}\)
Let P = 3k and H = 4k.
In right angled ΔABC,
P² + B² = H² (By Pythagoras theorem)
⇒ (3k)² + B² = (4k)²
⇒ 9k² + B² = 16k²
⇒ B² = 16k² – 9k² = 7k²
∴ B = +k\(\sqrt{7}\) (∵ Base cannot be -ve)

Question 4.
Given 15 cot A = 8, find sin A and sec A.
Solution:

Given, 15 cot A = 8 ⇒ cot A = \(\frac{8}{15}\)
\(\frac{\mathrm{B}}{\mathrm{P}}=\frac{8}{15}\)
B = 8k and P = 15 k
In right angled ΔABC, H² = B² + P² (By Pythagoras theorem)
= (8k)² + (15k)²
= 64k² + 225k²
= 289k²
∴ H = 17k (∵ Side cannot be -ve)
Now, sin A = \(\frac{\mathrm{P}}{\mathrm{H}}=\frac{15 \mathrm{k}}{17 \mathrm{k}}=\frac{15}{17}\)
sec A = \(\frac{\mathrm{H}}{\mathrm{B}}=\frac{17 \mathrm{k}}{8 \mathrm{k}}=\frac{17}{8}\)

Question 5.
Given sec θ = \(\frac{13}{12}\), calculate all other trigonometric ratios.
Solution :

sec θ = \(\frac{1}{cos θ}\)
sec θ . cos θ = 1
\(\frac{13}{12} \times \frac{12}{13}\) = 1
AB² = AC² – BC²
= 13² – 12²
= 169 – 144
= 25
AB = \(\sqrt{25}\) = 5

Question 6.
If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Solution :

cos A = \(\frac{AC}{AB}\)
cos B = \(\frac{BC}{AB}\)
cos A = cos B
\(\frac{AC}{AB}=\frac{BC}{AB}\)
∴ AC = BC
∴ \(\hat{A}\) = \(\hat{B}\) (∵ Angles opposite to equal sides are also equal)

Question 7.
If cot θ = \(\frac{7}{8}\), evaluate: (i) \(\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}\) (ii) cot² θ
Solution :

AC² = AB² + BC²
= 7² + 8²
= 49 + 64 = 113
AC = \(\sqrt{113}\)

(ii) cot² θ = (cot θ)²
= (\(\frac{7}{8}\))² = \(\frac{49}{64}\)
Given cot θ = \(\frac{7}{8}\)

Question 8.
If 3 cot A = 4, check whether \(\frac{1-\tan ^2 A}{1+\tan ^2 A}\) = cos² A – sin² A or not.
Solution :

3 cot A = 4
cot A = \(\frac{4}{3}\)
AC² = AB² + BC²
= 4² + 3²
= 16 + 9
= 25
AC = \(\sqrt{25}\) = 5

Question 9.
In triangle ABC, right-angled at B, if tan A = find the value of :
(i) sin A cos C + cos A sin C,
(ii) cos A cos C – sin A sin C.
Solution :
(i) sin A cos C + cos A sin C

(ii) cos A . cos C – sin A . sin C
\(\frac{AB}{AC} \cdot \frac{BC}{AC}\) – \(\frac{BC}{AC} \cdot \frac{AB}{AC}\)

Question 10.
In ΔPQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
Solution:
PR + QR = 25
QR = 25 – PR
PQ² + QR² = PR²
(5)² + (25 – PR)² = PR²
25 + (625 – 50PR + PR²) = PR²
650 – 50PR = PR² – PR² = 0
– 50PR = – 650
PR = \(\frac{-650}{-50}\) = 13.
PR = 13, QR = 25 – PR = 25 – 13 = 12, PQ = 5.

Question 11.
State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) sec A = \(\frac{12}{5}\) for some value of angle A.
(iii) cos A is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
(v) sin θ = \(\frac{4}{3}\) for some angle θ.
Solution :
(i) False, because the value of tan A increases from 0 to ∞. Also, tan 45° = 1.
(ii) True, because the value of sec A increases from 1 to ∞.
(iii) False, cos A is the abbreviation used for the cosine of angle A.
(iv) False, because cot A is one symbol. We cannot separate cot and A.
(v) False, because the value of sin θ always lies between 0 and 1. Here, sin θ = \(\frac{4}{3}\) which is greater than 1. So, it is not possible.