Question

in the given figure, AB is a diameter with centre and O and OP||BQ if angle ABQ= 40 , then the value of x is​

Answer 1

Answer:

70°

Step-by-step explanation:

⇒ ∠AOP = ∠OBQ = 40° (corresponding angles)

⇒ In ΔOAP, OA = OP = r (radius)

So ∠OAP = ∠OPA

Now, ∠OAP + ∠AOP + ∠OPA = 180°

⇒ 2∠OPA + 40° = 180° ⇒ ∠OPA = 70° (1)

ATQ, APQB is a cyclic quadrilateral.

So ∠APQ + ∠ABQ = 180° (opp. angles of cyclic quad.)

⇒ (∠APO + x) + 40° = 180°

(∵ ∠ABQ = ∠AOP)

⇒ 70° + x + 40° = 180°

⇒ x = 70°.

Answer 2

Answer:

70°

Step-by-step explanation:

angleB = angle O = 40 (corresponding angle)

In triangle AOP = OA=OP (radius of the circle)

angle A = angle P

Angle A + angle O +angle P =180

2.angle A + 40 = 180

2.angle A = 180-40

2.angle A = 140

angle A = 140/2

angle A = 70

angle P + angle B = 180

(angle P = 70 + x)

70 + x + 40 = 180

x + 110 = 180

x = 180 - 110

x = 70