In Fig. 16.199, AB is a diameter of the circle such that ∠A=35° and ∠Q=25°, find ∠PBR.
Answers
Given : AB is a diameter of the circle such that ∠A = 35° and ∠Q = 25°.
To Find : ∠PBR
Proof :
In ∆ ABQ,
Since Sum of the angles of a triangle is 180° :
∴ ∠ABQ + ∠AQB + ∠BAQ = 180°
∠ABQ + 25° + 35° = 180°
∠ABQ + 60° = 180°
∠ABQ = 180° - 60°
∠ABQ = 120°
∠ABQ + ∠ABR = 180° (Linear pair)
120° + ∠ABR = 180°
∠ABR = 180° - 120°
∠ABR = 60° ……….(1)
In ∆APB,
Since Sum of the angles of a triangle is 180° :
∠APB + ∠PBA + ∠PAB = 180°
Since angle in the semi-circle is 90° . i.e ∠APB = 90°
90° + ∠PBA + 35° = 180°
∠PBA + 125° = 180°
∠PBA = 180° - 125°
∠PBA = 55°
Now,
∠PBR = ∠PBA + ∠ABR
∠PBR = 55° + 60°
[From eq 1]
∠PBR = 115°
Hence, ∠PBR is 115°.
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Answer:-
Given :
AB is a diameter of the circle such that ∠A = 35° and ∠Q = 25°.
To Find :
∠PBR
Proof :
In ∆ ABQ,
Since Sum of the angles of a triangle is 180° :
∴ ∠ABQ + ∠AQB + ∠BAQ = 180°
∠ABQ + 25° + 35° = 180°
∠ABQ + 60° = 180°
∠ABQ = 180° - 60°
∠ABQ = 120°
∠ABQ + ∠ABR = 180° (Linear pair)
120° + ∠ABR = 180°
∠ABR = 180° - 120°
∠ABR = 60° ……….(1)
In ∆APB,
Since Sum of the angles of a triangle is 180° :
∠APB + ∠PBA + ∠PAB = 180°
Since angle in the semi-circle is 90° . i.e ∠APB = 90°
90° + ∠PBA + 35° = 180°
∠PBA + 125° = 180°
∠PBA = 180° - 125°
∠PBA = 55°
Now,
∠PBR = ∠PBA + ∠ABR
∠PBR = 55° + 60°
[From eq 1]
∠PBR = 115°
Hence, ∠PBR is 115°.
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