Question

AB is a diameter of a circle, with centre O. If ∠PAB = 55˚, ∠PBQ = 25˚, and ∠ABR = 50˚, find ∠PBA, ∠BPQ and ∠BAR.
Advertisement

Answer 1

According to question

Angle PAB = 55
Angle PBQ = 25
Angle ABR = 50

Thus
Angle PBA + Angle PAB + Angle BPA = 180
55 + 90 +  Angle PBA = 180
       Angle PBA = 180 - 145
                         = 35

Answer 2

Answer:

Step-by-step explanation:

Given AB is a diameter of a circle with centre O. If ∠PAB=55 ,∠PBQ= 25 and ∠ABR=50. We have to find the ∠PBA, ∠BPQ, ∠BAR

Now, ∠APB=∠ARB=90°   (∵angles in the semicircle)

In ΔAPB, By angle sum property of triangle

∠BAP+∠APB+∠PBA=180°

⇒ 55°+90°+∠PBA=180°

⇒ ∠PBA=35°

Now, ∠PQB=∠PAB=55°    (∵Angle subtended by the same chord)

In ΔPQB, By angle sum property of triangle

∠BPQ+∠PQB+∠QBP=180°

⇒ ∠BPQ+55°+25°=180°

⇒ ∠BPQ=100°

In ΔARB, By angle sum property of triangle

∠ARB+∠ABR+∠BAR=180°

⇒ 90°+50°+∠BAR=180°

⇒ ∠BAR=40°

I hope it was helpful...