Question

PQ is a tangent to a circle with centre O at the point P. If $$△OPQ$$ is an isosceles triangle with P as a vertex , then $$\mathrm{\angle }OQP$$ is equal to
a) $${30}^{\circ }$$
b) $${45}^{\circ }$$
c) $${60}^{\circ }$$
d) $${90}^{\circ }$$

Answer 1

Hint: We are given that PQ is a tangent to a circle with centre O at the point P. Let us consider OP to be the radius of the circle. Then, as PQ is tangent at point P, we know the tangent makes a right angle with the radius. Hence, $$\mathrm{\angle }OPQ={90}^{\circ }$$. Also, we are given that $$△OPQ$$ is an isosceles triangle. We will understand this using a diagram.

In the given figure, OP is the radius of the circle and PQ is tangent at point P. Hence, $$\mathrm{\angle }OPQ={90}^{\circ }$$. Since $$△OPQ$$ is an isosceles triangle, we have two sides of this triangle are equal. Let us consider $$OP=PQ$$. Now, we will use the property of an isosceles triangle that angles opposite to equal sides are equal. After that we will use the angle sum property of a triangle i.e. Sum of angles in a triangle is equal to $${180}^{\circ }$$ and find the required angle.

Complete step-by-step solution:
Since PQ is a tangent at point P and OP is the radius of the circle.
We see it through a figure.

We know, tangent makes a right angle with the radius and So,
$$\mathrm{\angle }OPQ={90}^{\circ }-----(1)$$
As $$△OPQ$$ is an isosceles triangle with $$OP=PQ$$.
We know, Angles opposite to equal sides are equal.
Angle opposite to $$PQ$$ is $$\mathrm{\angle }QOP$$.
Angle opposite to $$OP$$ is $$\mathrm{\angle }OQP$$.
Hence, $$\mathrm{\angle }OQP=\mathrm{\angle }QOP-----(2)$$
In $$△OPQ$$, using Angle Sum Property
$$\mathrm{\angle }OPQ+\mathrm{\angle }OQP+\mathrm{\angle }QOP={180}^{\circ }-----(3)$$
Using (1) and (2) in (3)
$${90}^{\circ }+\mathrm{\angle }OQP+\mathrm{\angle }OQP={180}^{\circ }$$
$$\Rightarrow {90}^{\circ }+2\mathrm{\angle }OQP={180}^{\circ }$$
Subtracting $${90}^{\circ }$$ both the sides, we get
$$\Rightarrow {90}^{\circ }+2\mathrm{\angle }OQP-{90}^{\circ }={180}^{\circ }-{90}^{\circ }$$
Clubbing the like terms on left hand side
$$\Rightarrow 2\mathrm{\angle }OQP+({90}^{\circ }-{90}^{\circ })={180}^{\circ }-{90}^{\circ }$$
Solving the left and the right hand side
$$\Rightarrow 2\mathrm{\angle }OQP+0^{\circ }={90}^{\circ }$$
$$\Rightarrow 2\mathrm{\angle }OQP={90}^{\circ }$$
Dividing both the sides by $$2$$.
$${\displaystyle \frac{2\mathrm{\angle }OQP}{2}}={\displaystyle \frac{{90}^{\circ }}{2}}$$
$$\mathrm{\angle }OQP={45}^{\circ }$$
Hence, we got $$\mathrm{\angle }OQP={45}^{\circ }$$.
Therefore, the correct option is (b).

Note: We need to be very thorough with the properties of circles and triangles. Once we interpret the given question, we need to think of the steps of solving the problem. While solving, we need to take care of the calculations. We have to read the question carefully and then draw the figure and then apply the properties. Each and every detail in the question needs to be read very carefully. While solving, we have to take care of each and every step.