Question

if PQ is a tangent to a circle at Q with center O,then angle OPQ is
A.45
B.90
C.120
D.100​

Answer 1

$\huge\underbrace{\overbrace{\boxed{\boxed{\mathfrak{Answer:}}}}}$

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°

 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°

⟹90°

+2∠OQP=180°

⟹∠OQP=45°

.

Ans- Option- A.

Answer 2

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°

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°

⟹90°

+2∠OQP=180°

⟹∠OQP=45°

.

Ans- Option- A.