In Fig. 16.200, P and Q are centres of two circles intersecting at B and C. ACD is a straight line. Then, ∠BQD =
Answers
Given : P and Q are centres of two circles intersecting at B and C. ACD is a straight line. ∠APB = 150°
To find = ∠BQD
Solution :
We know that, the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
∠ACB = ∠APB/2
∠ACB = 150°/2
∠ACB = 75°
Since, ACD is a straight line, so
∠ACB + ∠BCD = 180°
75o + ∠BCD = 180°
∠BCD = 180° – 75°
∠BCD = 105°
Now,
We know that, the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
∠BCD = ½ Reflex ∠BQD
105o = ½ (360° - ∠BQD)
105° × 2 = (360° - ∠BQD)
210° = 360° - ∠BQD
∠BQD = 360° – 210°
∠BQD = 150°
Hence, ∠BQD is 150°.
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Some questions of this chapter :
In Fig. 16.199, AB is a diameter of the circle such that ∠A=35° and ∠Q=25°, find ∠PBR.
https://brainly.in/question/15910508
In Fig. 16.198, A is the centre of the circle. ABCD is a parallelogram and CDE is a straight line. Find ∠BCD:∠ABE
https://brainly.in/question/15910499
Answer:
(Angle at the centre is double the angle at the circumference subtended by the same chord)