Jharkhand Board JAC Class 9 Maths Solutions Chapter 6 Lines and Angles Ex 6.3 Textbook Exercise Questions and Answers.

## JAC Board Class 9th Maths Solutions Chapter 6 Lines and Angles Ex 6.3

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Question 1.

In Fig, sides QP and RQ of ΔPQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.

Answer:

Given, ∠SPR = 135° and ∠PQT = 110°

∠SPR + ∠QPR = 180° (SQ is a straight line)

⇒ 135° + ∠QPR = 180°

⇒ ∠QPR = 45°

Also, ∠PQT +∠PQR = 180° (TR is a straight line)

⇒ 110° + ∠PQR= 180°

⇒ ∠PQR = 70°

Now, ∠PQR + ∠QPR + ∠PRQ = 180°

(Sum of the interior angles of the triangle)

⇒ 70° + 45° + ∠PRQ = 180°

⇒ 115° + ∠PRQ = 180°

⇒ ∠PRQ = 65°

Question 2.

In Fig, ∠X = 62°, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ΔXYZ, find ∠OZY and ∠YOZ.

Answer:

Given ∠X = 62°, ∠XYZ = 54°

YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively.

∠X + ∠XYZ + ∠XZY = 180° (Sum of the interior angles of the triangle)

⇒ 62° + 54° + ∠XZY = 180°

⇒ 116° + ∠XZY = 180°

⇒ ∠XZY = 64°

Now, ∠OZY = \(\frac {1}{2}\) ∠XZY (ZO bisector.)

⇒ ∠OZY = 32°

also, ∠OYZ = \(\frac {1}{2}\) ∠XYZ (YO is the bisector.)

⇒ ∠OYZ = 27°

Now, ∠OZY + ∠OYZ + ∠YOZ = 180° (Sum of the interior angles of the triangle)

⇒ 32° + 27° + ∠YOZ = 180°

⇒ 59° + ∠YOZ = 180°

⇒ ∠YOZ = 121°

Question 3.

In, Fig , if AB || DE, ∠BAC = 35° and ∠CDE = 53° and ∠DCE.

Answer:

Given, AB || DE, ∠BAC = 35° and ∠CDE = 53°

∠BAC = ∠CED (Alternate interior angles.)

∴ ∠CED = 35°

Now, ∠DCE + ∠CED + ∠CDE = 180° (Sum of the interior angles of the triangle)

⇒ ∠DCE + 35° + 53° = 180°

⇒ ∠DCE + 88° = 180°

⇒ ∠DCE = 92°

Question 4.

In Fig, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.

Answer:

Given, ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°

∠PRT + ∠RPT + ∠PTR = 180° (Sum of the interior angles of the triangle)

⇒ 40° + 95° + ∠PTR = 180°

⇒ 135° + ∠PTR = 180°

⇒ ∠PTR = 45°

∠STQ = ∠PTR = 45° (Vertically opposite angles)

Now, ∠TSQ + ∠STQ + ∠SQT = 180° (Sum of the interior angles of the triangle)

⇒ 75° + 45° + ∠SQT = 180°

⇒ 120° + ∠SQT = 180°

⇒ ∠SQT = 60°

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Question 5.

In Fig, if PQ || PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.

Answer:

Given, PQ || PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°

x + ∠SQR = ∠QRT (Alternate angles as QR is transversal)

⇒ x + 28° = 65°

⇒ x = 37°

also, ∠QSR = x (Alternate angles)

⇒ ∠QSR = 37°

also, ∠QRS+∠QRT=180°(Linear pair)

⇒ ∠QRS + 65° = 180°

⇒ ∠QRS =115°

Now, ∠P + ∠Q+ ∠R + ∠S = 360°

(Sum of the angles in a quadrilateral)

⇒ 90° + 65° + 115° + ∠S = 360°

⇒ 270° + y + ∠QSR = 360°

⇒ 270° + y + 37° = 360°

⇒ 307° + y = 360°

⇒ y = 53°

Question 6.

In Fig, the side QR of APQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = \(\frac{1}{2}\) ∠QPR.

Answer:

Given: Bisectors of ∠PQR and ∠PRS meet at point T.

To prove: ∠QTR = \(\frac{1}{2}\) ∠QPR.

Proof: ∠TRS = ∠TQR + ∠QTR (Exterior angle of a triangle equals the sum of its interior opposite angles.)

⇒ ∠QTR = ∠TRS – ∠TQR …………(i) also,

∠SRP = ∠QPR + ∠PQR

∵ 2 ∠TRS = ∠QPR + 2 ∠TQR

⇒ ∠SRP = 2∠TRS and ∠PQR = 2∠TQR

⇒ ∠QPR = 2 ∠TRS – 2 ∠TQR

⇒ \(\frac{1}{2}\) QPR = ∠TRS – ∠TQR …………(ii)

Equating (i) and (ii)

∠QTR = \(\frac{1}{2}\) ∠QPR

∴ Hence, proved.