Question

PQ is a tangent to a circle with centre O at point P. If AOPQ is an isosceles
triangle, then find OQP.​

Answer 1

Step-by-step explanation:

Given- O is the centre of a circle to which PQ is a tangent at P. ΔOPQ is isosceles whose vertex is P.

To find out- ∠OQP=?

Solution- OP is a radius through P, the point of contact of the tangent PQ with the given circle ∠OPQ=90

o

since the radius through the point of contact of a tangent to a circle is perpendicular to the tangent. Now ΔOPQ is isosceles whose vertex is P.

∴OP=PQ⟹∠OQP=∠QOP⟹∠OQP+∠QOP=2∠OQP.

∴ By angle sum property of triangles,

∴∠OPQ+2∠OQP=180

o

⟹90

o

+2∠OQP=180

o

⟹∠OQP=45

o

.

Answer 2

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pq is tangent and Oq is radius

we know that tangent is perpendiculatr to radius

Angle Opq = 90°

Angle OPQ is isoceles

• OP = PQ

Angle OQP = Angle POQ

SO,

angle Oqp = 1/2 X 90°

= 45°

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