Question

The angle between two tangents drawn from an external point to a circle with centre O is 50°. If the tangents meet the circle at P & Q, then angle POQ is equal to__________​

Answer 1

Answer:

Ap=BP ....(length of tangents from external point to circle are equal)

∠A=∠B=90

o

.... (Tangent is ⊥ to radius)

OP=OP .... (common side)

∴△AOP≅△BOP ....(RHS test of congruence)

∠APO=∠BPO=30

o

→c.a.c.t

∠AOP=∠BOP=60

o

→c.a.c.t

△AOP is 30

o

−60

o

−90

o

triangle.

∴△AOP,

cos60=

OP

OA

OP=

2

1

a

=2a

solution

Answer 2

Step-by-step explanation:

Given:The angle between two tangents drawn from an external point to a circle with centre O is 50°. If the tangents meet the circle at P & Q.

To find: Angle POQ is equal to__________.

Solution:

Tip: Tangent form 90° with radius, where it touches the circle.

Let the external point is R.

$\angle PRQ=50°$[Given]

$\angle OPR=90°$[From theorem]

$\angle OQR=90°$[From theorem]

because, POQR is a quadrilateral, so

Sum of internal angles should be 360°.

$\angle PRQ+\angle OPR+\angle OQR+\angle POQ=360°$

$50°+90°+90°+\angle POQ=360°$

$\angle POQ=360°-230°$

$\bf \angle POQ=130°$

Final answer:

$\bf \angle POQ=130°$

Hope it will help you.

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