Question

PQ is a tangent to a circle with centre 0 at the point P. If A Δ OPQ is an isosceles triangle, then ∠OQP is equal to
A. 30°
B. 45°
C. 60°
D. 90°

Answer 1

Given : PQ is a tangent to a circle with centre O at the point P and  Δ OPQ is an isosceles triangle.

 

To find : ∠OQP is equal to

 

Solution :  

We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact. Then,

∆POQ is right-angled at ∠OPQ (i.e., ∠OPQ = 90°).

∵ ∆POQ is an isosceles triangle

∴ ∠POQ = ∠OQP

∠POQ + ∠OQP + ∠OPQ = 180°

[Sum of all angles of a triangle is180°]

∠OQP + ∠OQP + ∠OPQ = 180°

2∠OQP + 90° = 180°

2∠OQP = 180° - 90°

2∠OQP = 90°

∠OQP = 90°/2

∠OQP = 45°

Hence, ∠OQP is 45°.

Among the given options option (B) 45° is correct.

HOPE THIS ANSWER WILL HELP YOU……

 

Some more questions :  

In Fig. 10.58, there are two concentric circles with centre O of radii 5 cm and 3 cm. from an external point P, tangents PA and PB are drawn to these circles. If AP = 12 cm, find the length of BP.

https://brainly.in/question/15932446

 

TP and TQ are the tangents from the external point T of a circle with centre O. If ∠ OPQ = 30° then find the measure of ∠ TQP.

brainly.in/question/3130906

Answer 2

GIVEN :-

• PQ is a tangent to a circle with centre O at the point P.

• Δ OPQ is an isosceles triangle.

TO FIND :-

The value of ∠OQP(The angle is given as θ in the attachment).

HERE'S WHAT WE NEED TO KNOW :-

→ The radius to the tangent at the point of contact are perpendicular to each other.

→ So, Δ OPS is a right-isosceles triangle.

→ Sum of all the angles of a triangle = 180°(By Angle Sum Property)

SOLUTION :-

We know,

∠OQP + ∠POQ  + ∠OPQ = 180°

⇒ ∠OQP + ∠OQP + ∠OPQ = 180°   (As ∠OQP = ∠POQ)

⇒ 2∠OQP = 180° - ∠OPQ

⇒ 2∠OQP = 180° - 90°                    (As it is perpendicular at that point)

⇒ 2∠OQP = 90°

⇒ ∠OQP = 90°/2

⇒ ∠OQP = 45°

So, we conclude that, the value of ∠OQP is 45°.

The answer is B. 45°