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.
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°.
Related Questions
In the given figure, A is the center of the circle, ABCD is a parallelogram and CDE is a straight line. Prove that : ∠BCD = 2∠ABE.
In the given figure, I is the incenter of △ABC, BI when produced meets the circumcircle of △ABC at D. Given, ∠BAC = 55° and ∠ACB = 65°; calculate :
(i) ∠DCA,
(ii) ∠DAC,
(iii) ∠DCI,
(iv) ∠AIC.
In the figure given alongside, AB || CD and O is the center of the circle. If ∠ADC = 25°; find the angle AEB. Give reasons in support of your answer.
In the figure given alongside, AB and CD are straight lines through the centre O of a circle. If ∠AOC = 80° and ∠CDE = 40°, find the number of degrees in :
(i) ∠DCE,
(ii) ∠ABC.
