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.