30_2_1_Mathematics Std._cl.10

[CBSE Class 10 2026] :Finding the Angle Between Tangents , PA & PB are tangents to a circle centred at O, If…

Here is a classic geometry problem involving a circle and its tangents. We have a circle with center O, and two tangents, PA and PB, drawn from an external point P. We're given that angle OAB is 15 degrees, and we need to find angle APB.

Let's focus on the angle we know. The line segment OA is a radius, and so is OB. Because all radii of a circle are equal, triangle OAB is an isosceles triangle.

Now, let's recall a fundamental property of tangents. A tangent to a circle is always perpendicular to the radius at the point of contact. This means the entire angle OAP is exactly 90 degrees.

We can use this to find the angle inside the triangle PAB. Since the whole angle OAP is 90 degrees, and the small piece OAB is 15 degrees, the remaining part—angle PAB—must be 90 minus 15, which is 75 degrees.

Here's where it gets interesting. Another key property is that two tangents drawn from the same external point are always equal in length. So, PA equals PB. This makes triangle PAB an isosceles triangle too! So angle PBA is also 75 degrees.

Finally, we can find our missing angle. The sum of angles in any triangle is 180 degrees. In triangle PAB, we have two 75-degree angles, which add up to 150. That leaves exactly 30 degrees for angle APB. And that's our answer!