Solveeit Logo
Login

Question

Question: Given ∆ABC is inscribed in the semicircle with diameter AB. The area of ∆ABC equals 2/9 of the area ...

Given ∆ABC is inscribed in the semicircle with diameter AB. The area of ∆ABC equals 2/9 of the area of the semicircle. If the measure of the smallest angle in ∆ABC is x, then sin2x is equal to –

A

π9\frac { \pi } { 9 }

B

2π9\frac { 2 \pi } { 9 }

C

π18\frac { \pi } { 18 }

D

π8\frac { \pi } { 8 }

Answer

π9\frac { \pi } { 9 }

Explanation

Solution

Since, a2 + b2 = c2 = 4r2 …..(i)

Also, 12\frac { 1 } { 2 }a.b =

⇒ 9ab = 2πr2 ..…(ii)

From Eqs. (i) and (ii), we get

= 2π\frac { 2 } { \pi }⇒ + = 18π\frac { 18 } { \pi }

Let ∠ BAC, then cos⁡xsin⁡x\frac { \cos x } { \sin x }+ sin⁡xcos⁡x\frac { \sin x } { \cos x }= 18π\frac { 18 } { \pi }

⇒= 18π\frac { 18 } { \pi }

⇒sin x . cos x = π18\frac { \pi } { 18 }

⇒ sin 2x = π9\frac { \pi } { 9 }