Haryana State Board HBSE 9th Class Maths Solutions Chapter 6 Lines and Angles Ex 6.3 Textbook Exercise Questions and Answers.

## Haryana Board 9th Class Maths Solutions Chapter 6 Lines and Angles Exercise 6.3

Question 1.

In figure 6.79, sides QP and RQ of ΔPQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.

Solution:

Since, TQR is a straight line.

∴ ∠PQT + ∠PQR = 180°,

(Linear pair axiom)

⇒ 110° + ∠PQR = 180°

⇒ ∠PQR = 180° – 110°

⇒ ∠PQR = 70° …(i)

In ΔPQR, we have

∠SPR = ∠PQR + ∠PRQ,

[By theorem 6.8]

⇒ 135° = 70° + ∠PRQ,

[∵ ∠SPR = 135 and using (i)]

⇒ 135° – 70° = ∠PRQ

⇒ ∠PRQ = 65°

Hence, ∠PRQ = 65°.

Question 2.

In figure 6.80, ∠X = 62°, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ΔXYZ, find ∠OZY and ∠YOZ.

Solution:

In ΔXYZ, we have

∠X + ∠Y + ∠Z = 180°,

[∵ Sum of angles of a triangle = 180°

⇒ 62° + 54° + ∠Z = 180°,

(∵ ∠X = 62° and ∠Y = 54°)

⇒ 116° + ∠Z = 180°

⇒ ∠Z = 180° – 116° = 64°

Now, ∠OZY = \(\frac {1}{2}\)∠XZY [∵ ZO is the bisector of ∠XZY]

⇒ ∠OZY = \(\frac {64°}{2}\)

⇒ ∠OZY = 32° …….(i)

and ∠OYZ = \(\frac {1}{2}\)∠XYZ (∵ YO is the bisector of ∠XYZ)

⇒ ∠OYZ = \(\frac {54°}{2}\)

⇒ ∠OYZ = 27° …….(ii)

Now, in ΔYOZ, we have

∠OYZ + ∠OZY + ∠YOZ = 180°,

(Sum of angles of a triangle = 180°)

⇒ 27° + 32° + ∠YOZ = 180°,

[Using (i) and (ii)]

⇒ 59° + ∠YOZ = 180°

⇒ ∠YOZ = 180° – 59°

⇒ ∠YOZ = 121°

Hence, ∠OZY = 32° and ∠YOZ = 121°.

Question 3.

In figure 6.81, if AB || DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.

Solution :

∵ AB || DE and AE is the transversal

∴ ∠AED = ∠BAE,

(Alternate interior angles)

⇒ ∠AED = 35°

⇒ ∠CED = 35°,

[∵ ∠AED = ∠CED]

In ΔCDE, we have ∠CDE + ∠DCE + ∠CED = 180°

(∵ Sum of angles of a triangle = 180°)

⇒ 53° + ∠DCE + 35° = 180°

⇒ 88° + ∠DCE = 180°

⇒ ∠DCE = 180° – 88°

⇒ ∠DCE = 92°

Hence, ∠DCE = 92°

Question 4.

In figure 6.82, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95o and ∠TSQ = 75°, find ∠SQT.

Solution:

In ΔPRT, we have

∠RPT + ∠PRT + ∠PTR = 180°,

(∵ Sum of angles of a triangle = 180°)

⇒ 95° + 40° + ∠PTR = 180°

⇒ 135° + ∠PTR = 180°

⇒ ∠PTR = 180° – 135°

⇒ ∠PTR = 45°

Now, ∠STQ = ∠PTR,

(Vertically opposite angles)

⇒ ∠STQ = 45°

Now in triangle ΔTSQ, we have

∠STQ + ∠TSQ + ∠SQT = 180°,

(∵ Sum of angles of a triangle = 180°)

⇒ 45° + 75° + ∠SQT = 180°

⇒ 120° + ∠SQT = 180°

⇒ ∠SQT = 180° – 120°

⇒ ∠SQT = 60°

Hence, ∠SQT = 60

Question 5.

In figure 6.83, if PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.

Solution :

Since, PQ || SR and QR is the transversal

∴ ∠PQR = ∠QRT,

(Alternate interior angles)

⇒ ∠PQR = 65°

⇒ x + 28° = 65°,

(∵ ∠PQR = x + 28°)

⇒ x = 65° – 28°

⇒ x = 37°

Now, in ΔPSQ, we have

∠SPQ + ∠PSQ + ∠PQS = 180°,

(Sum of angles of a triangle = 180°)

⇒ 90° + y + x = 180°

⇒ 90° + y + 37° = 180°

⇒ 127° + y = 180°

⇒ y = 180° – 127°

⇒ y = 53°

Hence, x= 37° and y = 53°.

Question 6.

In figure 6.84, the side QR of ΔPQR 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.

Solution:

In ΔPQR, we have

∠PRS = ∠QPR + ∠PQR, (By theorem 6.8)

⇒ \(\frac {1}{2}\)∠PRS = \(\frac {1}{2}\)∠QPR + \(\frac {1}{2}\)∠PQR

⇒ ∠TRS = \(\frac {1}{2}\)∠QPR + ∠TQR …….(i)

[∵ RT and QT are the bisectors of ∠PRS and ∠PQR respectively

∴ ∠TRS = \(\frac {1}{2}\)∠PRS and ∠TQR = \(\frac {1}{2}\)∠PQR]

In ΔTQR, we have

∠TRS = ∠QTR + ∠TQR …(ii)

(By theorem 6.8)

From (i) and (ii), we get

\(\frac {1}{2}\)∠QPR + ∠TQR = ∠QTR + ∠TQR

⇒ \(\frac {1}{2}\)∠QPR = ∠QTR

⇒ ∠QTR = \(\frac {1}{2}\)∠QPR.

Hence proved