Question

The perimeter of ΔABC is 20, ∠A = 60°, area of ΔABC = `10sqrt(3)`, then find the values of a, b, c.

Answer 1

a + b + c = 20, 

A(ΔABC) = `10sqrt(3)`

∠A = 60°

Area of triangle = `1/2 bc sin A`

∴ `10sqrt(3) = 1/2 bc * sqrt(3)/2`

∴ bc = 40

Also cos A = `(b^2 + c^2 - a^2)/(2bc)`

i.e. a² = b² + c² – 2bc cos A

i.e. a² = `b^2 + c^2 - 2(40)(1/2)`

∴ a² = b² + c² – 40

∴ a² = (b² + c² + 2bc) – 2bc – 40

∴ (b + c)² – 2(40) – 40

∴ (20 – a)² – 120   ...(∵ a + b + c = 20)

∴ a² = 400 – 40a + a² – 120

∴ 40a = 280,

∴ a = 7

But a + b + c = 20

∴ b + c = 13 and bc = 40

∴ b(13 – b) = 40

∴ b² – 13b + 40 = 0

∴ (b – 8)(b – 5) = 0 

∴ b = 8 or b = 5

From bc = 40

When b = 8 `\implies` c = 5 and

When b = 5 `\implies` c = 8

∴ a, b, c are 7, 8, 5 or 7, 5, 8 respectively.