Question

In Fig. 16.194, two congruent circles with centres O and O’ intersect at A and B. If ∠AO’B=50°, then find ∠APB.

Answer 1

Answer:

130

Step-by-step explanation:

angle AOB=50

angleAPB=?

APB+AOB=180

APB+50=180

APB=130

Answer 2

Given:

Angle AO'B=50°

To find:

The angle APB

Solution:

The measure of angle APB is 25°.

We can find the angle by following the given steps-

We know that the two circles are congruent.

So, the radii of the circles with centre O and O' are equal.

We are given that the angle AO'B=50°

The chord AB is common in both circles.

So, angle AO'B=angle AOB

Angle AOB=50°

Now, POB is a straight line and the angles POA and AOB form a linear pair.

The sum of angle POA and angle AOB=180°

angle POA +angle AOB=180°

Angle POA+50°=180°

Angle POA=180°-50°

Angle POA=130°

In triangle POA, OP and OP are equal. (Radii of the circle)

So, angle APB=angle OAP. (Angles opposite to equal sides are equal)

Also, angle APB+angle OAP+angle POA=180° (Sum of angles of a triangle)

2×angle APB+130°=180°

2×angle APB=180°-130°

2×angle APB=50°

Angle APB=50°/2

Angle APB=25°

Therefore, the measure of angle APB is 25°.