solve the following
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(iii) ∠ PAB = 90°
in Δ PAB
∠PAB+∠APB+∠PBA=180°
∠PBA=30°
(i)join OA
In Δ OBA
let ∡ BOA be θ°
OB=OA [∵radius]
∠OBA= ∠OAB = 30°
∠OBA+∠OAB+∠BOA= 180°
θ= 120°
m (arc AQB) = θ/360 × 2πr
= 2πr/3
(ii) PB ⏊ BC [∵BC is a tangent]
∠ABC = 180°-∠PBA
= 60°
(iv) m(arc APB) = 2πr- m(arcAQB)
= 2πr-2πr/3
= 4πr/3
in Δ PAB
∠PAB+∠APB+∠PBA=180°
∠PBA=30°
(i)join OA
In Δ OBA
let ∡ BOA be θ°
OB=OA [∵radius]
∠OBA= ∠OAB = 30°
∠OBA+∠OAB+∠BOA= 180°
θ= 120°
m (arc AQB) = θ/360 × 2πr
= 2πr/3
(ii) PB ⏊ BC [∵BC is a tangent]
∠ABC = 180°-∠PBA
= 60°
(iv) m(arc APB) = 2πr- m(arcAQB)
= 2πr-2πr/3
= 4πr/3
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