Question

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

Answer 1

Step-by-step explanation:

I hope it will be help full for you

Answer 2

Given :-

A circle with centre O such that

• PQ is a tangent to a circle at P.

• ∆ OPQ is isosceles.

To Find :-

• ∠ OQP

Properties Used :-

• 1. Radius and tangent are perpendicular to each other.

• 2. In right angle triangle, Hypotenuse is the longest side.

• 3. Angle opposite to equal sides are always equal.

• 4. Sum of all interior angles of a triangle is 180°.

$\large\underline{\sf{Solution-}}$

Since,

PQ is a tangent to a circle with cente O at point P and OP is radius of circle,

$\bf\implies \:OP \: \perp \: PQ$

$\rm :\implies\: \angle \: OPQ \: = \: 90 \degree$

$\rm :\implies\: \triangle \: OPQ \: is \:right \: angle \: \triangle$

Now,

it is given that

• ∆OPQ is isosceles.

$\rm :\implies\:OP \: = \: PQ$

$\rm :\implies\: \angle \: PQO \: = \angle \: QOP \:$

Now,

• In ∆ OPQ,

We know that,

• Sum of all interior angles of a triangle is 180°.

So,

$\rm :\longmapsto\:\angle \: QOP \: + \angle \: PQO + \angle \: OPQ = 180 \degree$

$\rm :\longmapsto\:\angle \: PQO + \angle \: PQO + 90 \degree \: = \: 180\degree$

$\rm :\longmapsto\:2\angle \: PQO = 180\degree \: - \: 90\degree$

$\rm :\longmapsto\:2\angle \: PQO = \: 90\degree$

$\rm :\longmapsto\:\angle \: PQO = \: 45\degree$

$\bf\implies \: \angle \: OQP \: = \: 45\degree$

Additional Information :-

• 1. Length of tangents drawn from external point are equal.

• 2. Tangents are equally inclined to the line segment joining centre and external point.

• 3. One and only one tangent can be drawn at a point on circle.

• 4. From external point, two tangents can be drawn to a circle.