# Video: CBSE Class X • Pack 5 • 2014 • Question 12

CBSE Class X • Pack 5 • 2014 • Question 12

03:13

### Video Transcript

𝑃𝑄 and 𝑃𝑅 are two tangents to a circle with centre 𝑂 such that the angle 𝑄𝑃𝑅 is equal to 120 degrees, and point 𝑃 is external to the circle. Prove that two 𝑃𝑄 is equal to 𝑃𝑂.

Let’s begin by sketching this out. Remember our diagram does not need to be to scale. But it should be roughly in proportion to help us decide how best to answer the question.

First, we can add in the lines joining the centre of the circle 𝑂 to points on its circumference at 𝑄 and 𝑅. By their very definition, these are the radii of the circle. This means they’re of equal length.

Next, we should recall some of the circle theorems that might help here. The first circle theorem we can use says that a radius and a tangent are perpendicular. This means that the angles formed at 𝑂𝑄𝑃 and 𝑂𝑅𝑃 are both 90 degrees. We also know that the tangents drawn from an external point to a circle are of equal length. So we can see that the line segments joining 𝑃 and 𝑄 and 𝑃𝑅 must be equal.

This means that the two triangles 𝑂𝑄𝑃 and 𝑂𝑅𝑃 are congruent. They each have a right angle and two sides of equal length given by the tangent and the radii. Alternatively, we could have noticed that the line 𝑂𝑃 is a shared side to the triangle. And we could’ve used the RHS condition that says that each triangle has a right angle, a hypotenuse in common, and another side of equal length.

We are given that angle 𝑄𝑃𝑅 is 120 degrees. Since the triangles are congruent, this means that we can divide 120 exactly in two to get the measure of the angle at 𝑄𝑃𝑂. 120 divided by two is 60 degrees. Now, we have a right-angled triangle 𝑂𝑄𝑃, for which we know the measure of one of its angles. And we’re looking to form a relationship between the sides 𝑃𝑄 and 𝑃𝑂.

We’ll use right angle trigonometry to do so. 𝑃𝑄 is the side next to the included angle; that’s the adjacent. 𝑂𝑃 is the hypotenuse; that’s the one opposite the right angle. We will, therefore, use the cosine ratio, where cos of 𝜃 is equal to adjacent divided by hypotenuse. In this case, 𝜃 is 60 degrees. So this becomes cos of 60 is equal to the length of 𝑃𝑄 divided by the length of 𝑃𝑂.

Remember though we know that cos of 60 is one-half. So we can rewrite this as a half is equal to 𝑃𝑄 over 𝑃𝑂. And we can begin to rearrange this to get an equation in terms of 𝑃𝑂 by multiplying both sides by 𝑃𝑂. That gives us a half of 𝑃𝑂 is equal to 𝑃𝑄.

Our last step is to multiply everything by two. And doing so, we can see that 𝑃𝑂 is equal to two 𝑃𝑄 as required.