AB is a diameter of a circle with centre O. If angle PAB=55 ,angle PBQ= 25 and angle ABR=50,find angle PBA,angle BPQ angle BAR
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<pba=35°,<bpq=30°,<bar=40°
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Answer:
∠PBA=35°, ∠BPQ=100°, ∠BAR=40°
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°
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