Mathematics
AB is a diameter of the circle APBR as shown in the figure. APQ and RBQ are straight lines. Find :
(i) ∠PRB,
(ii) ∠PBR,
(iii) ∠BPR.
Circles
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Answer
(i) We know that,
Angles in same segment are equal.
∠PRB = ∠PAB = 35°.
Hence, ∠PRB = 35°.
(ii) From figure,
∠BPA = 90° [Angle in semi-circle is a right angle.]
⇒ ∠BPA + ∠BPQ = 180° [Linear pairs]
⇒ 90° + ∠BPQ = 180°
⇒ ∠BPQ = 180° - 90° = 90°.
Exterior angle of a triangle is equal to the sum of two opposite interior angles.
⇒ ∠PBR = ∠BPQ + ∠BQP = 90° + 25° = 115°.
Hence, ∠PBR = 115°.
(iii) In △ABP,
⇒ ∠BPA + ∠PAB + ∠ABP = 180°
⇒ 90° + 35° + ∠ABP = 180°
⇒ 125° + ∠ABP = 180°
⇒ ∠ABP = 180° - 125° = 55°.
From figure,
∠ABR = ∠PBR - ∠ABP = 115° - 55° = 60°.
∠APR = ∠ABR = 60°. [Angles in same segment are equal]
In △BPR,
⇒ ∠BPR + ∠PRB + ∠PBR = 180°
⇒ ∠BPR + 35° + 115° = 180°
⇒ 150° + ∠BPR = 180°
⇒ ∠BPR = 180° - 150° = 30°.
Hence, ∠BPR = 30°.
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