Question

In the given figure, AB is a diameter of the circle such that ∠A = 35° and ∠Q = 25°, find ∠PBR.

Answer 1

Let us first consider the triangle ΔABQ.

It is known that in a triangle the sum of all the interior angles add up to 180°.

So here in our triangle ΔABQ we have,

${\displaystyle \angle BAQ+\angle AQB+\angle ABQ}$ = 180°

                                  ${\displaystyle \angle ABQ=180°-\angle BAQ-\angle AQB}$

                                                = 180° - 35° - 25° 

                                  ${\displaystyle \angle ABQ}$  = 120° 

By a property of the circle we know that an angle formed in a semi-circle will be 90°..

In the given circle since ‘AB’ is the diameter of the circle the angle ${\displaystyle \angle APB}$  which is formed in a semi-circle will have to be 90°.

So, we have  ${\displaystyle \angle APB}$ = 90°

Now considering the triangle  ΔAPB  we have,

${\displaystyle \angle APB+\angle BAP+\angle ABP}$ = 180°

                                 ${\displaystyle \angle ABP =180°-\angle APB-\angle BAP}$

                                               = 180° - 90° - 35 °

                                  ${\displaystyle \angle ABP}$  = 55°

From the given figure it can be seen that,

${\displaystyle \angle ABP+\angle PBQ=\angle ABQ}$

              ${\displaystyle \angle PBQ=\angle ABQ-\angle ABP}$

                            = 120°- 55°

                ${\displaystyle \angle PBQ}$ = 65°

Now, we can also say that,

${\displaystyle \angle PBQ+\angle PBR}$ = 180°

                   ${\displaystyle \angle PBR=180°-\angle PBQ}$

                                 =180° - 65 °

                   ${\displaystyle \angle PBR}$  = 115°

Hence the measure of the angle  ${\displaystyle \angle PBR}$  is 115°.