Question

(1) O is the centre and PB is a
diameter of the circle. BC is the tangant of the circle with B as point of contact. If Z APB = 60°
then (i) m(arc AQB) = ?
(ii) Z ABC = ?
(iii) Z PBA = ?
(iv) m(arc APB) = ?
​

Answer 1

Answer:

Hey Buddy!!!

Given O is the center and PB is a diameter of the circle. BC is the tangent of the circle with B as point of contact. If angle APB= 60° then we have to find the following measures

By theorem,  The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.

∴ m(arc AQB)=2∠APB=2(60)=120°

Since ∠PAB is the angle in semicircle ∴ ∠PAB=90°

In ΔABP, by angle sum property of triangle

∠BPA+∠PAB+∠ABP=180°

⇒ 60°+90°+∠PBA=180° ⇒ ∠PBA=30°

∠PBC=∠PBA+∠ABC

⇒ 90°=30°+∠ABC ⇒ ∠ABC=60°

m(arc APB)=360°-m(arc AQB)

               = 360°-120°=240°

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Tc!!

Hope it helps...