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Question: If the radius of the circum-circle of an isosceles ∆ABC is equal to AB (= AC), then A! is-...

If the radius of the circum-circle of an isosceles ∆ABC is equal to AB (= AC), then A! is-

A

π4\frac { \pi } { 4 }

B

2π3\frac { 2 \pi } { 3 }

C

π3\frac { \pi } { 3 }

D

π2\frac { \pi } { 2 }

Answer

2π3\frac { 2 \pi } { 3 }

Explanation

Solution

= bsin⁡B\frac { b } { \sin B } = asin⁡A\frac { a } { \sin A } = 2R

⇒ bsin⁡θ\frac { \mathrm { b } } { \sin \theta } = = asin⁡(180−2θ)\frac { a } { \sin ( 180 - 2 \theta ) }= 2b (Q B! = C! = θ , say)

⇒ = asin⁡2θ\frac { a } { \sin 2 \theta } = 2b ⇒ sin θ = b2b\frac { b } { 2 b } ⇒ sin θ = 12\frac { 1 } { 2 }⇒ θ = π6\frac { \pi } { 6 }

∴ A! = π – 2θ = 2π3\frac { 2 \pi } { 3 }