Question

In a ΔABC, if cos A cos B cos C = `(sqrt(3) - 1)/8` and sin A sin B sin C = `(3 + sqrt(3))/8`, then the angles of the triangle are ______.

Options

• 45°, 60° and 75°

• 30°, 60° and 75°

• 45°, 30° and 75°

• 75°, 90° and 30°

Answer 1

In a ΔABC, if cos A cos B cos C = `(sqrt(3) - 1)/8` and sin A sin B sin C = `(3 + sqrt(3))/8`, then the angles of the triangle are 45°, 60° and 75°.

Explanation:

We have, cos A cos B cos C = `(sqrt(3) - 1)/8`

and sin B sin C = `(3 + sqrt(3))/8`

= tan A tan B tan C = `(3 + sqrt(3))/(sqrt(3) - 1)`  ...(i)

But tan A + tan B + tan C = tan A tan B tan C

`\implies` tan A + tan B + tan C = `(3 + sqrt(3))/(sqrt(3) - 1)`  ...(ii)

Now, A + B + C = π `\implies` cos(A + B + C) = cos π

`\implies` cos A cos B cos C – cos A sin B sin C – cos B sin C sin A – cos C sin A sin B = – 1

`\implies` cos A cos B cos C[1 – tan B tan C – tan C tan A – tan A tan B] = – 1

`\implies` `(sqrt(3) - 1)/8{1 - tan A tan B - tan B tan C - tan C tan A}` = – 1

`\implies` tan A tan B + tan B tan C + tan C tan A = `1 + 8/(sqrt(3) - 1)`

`\implies` tan A tan B + tan B tan C + tan C tan A = `5 + 4sqrt(3)`  ...(iii)

From equations (i), (ii), (iii), we find that tan A, tan B and tan C are the roots of the equation

`x^3 - ((3 + sqrt(3))/(sqrt(3) - 1))x^2 + (5 + 4sqrt(3))x - ((3 + sqrt(3))/(sqrt(3) - 1))` = 0

or `x^3 - (2 + sqrt(3))sqrt(3)x^2 + (5 + 4sqrt(3))x - (2 + sqrt(3))sqrt(3)` = 0

or `x^3 - (3 + 2sqrt(3))x^2 + (5 + 4sqrt(3))x - (3 + 2sqrt(3))` = 0

Clearly, x = 1 is a root of the above equation.

So, the given equation may be written as

`(x - 1){x^2 - (2 + 2sqrt(3))x + (3 + 2sqrt(3))}` = 0

`\implies (x - 1)(x - sqrt(3)){x - (2 + sqrt(3))}` = 0

`\implies x = 1, sqrt(3), 2 + sqrt(3)`

`\implies` tan A = 1, tan B = `sqrt(3)` and tan C = `2 + sqrt(3) = (sqrt(3) + 1)/(sqrt(3) - 1)`

`\implies` A = 45°, B = 60° and C = 75°

Hence, the angles of the triangle are 45°, 60° and 75°.