Question

In a ΔPQR, if 3 sin P + 4 cos Q = 6 and 4 sin Q + 3 cos P = 1, then ∠R is equal to ______.

Options

• `(5π)/6`

• `π/6`

• `π/4`

• `(3π)/4`

Answer 1

In a ΔPQR, if 3 sin P + 4 cos Q = 6 and 4 sin Q + 3 cos P = 1, then ∠R is equal to `underlinebb(π/6)`.

Explanation:

Given, in ΔPQR,

3 sin P + 4 cos Q = 6   ...(i)

and 4 sin Q + 3 cos P = 1  ...(ii)

On squaring and adding the equations (i) and (ii), we get

9 (sin² P + cos² P) + 16 (sin² Q + cos² Q) + 2 × 3 × 4 (sin P cos Q + sin Q cos P) = 37

`\implies` 24 [sin (P + Q)] = 37 – 25

`\implies` sin (P + Q) = `1/2`

Since, P and Q are angles of ΔPQR.

0° < P, Q < 180°

`\implies` P + Q = 30° or 150°

`\implies` R = 150° or 30°   ...[respectively]

Hence, two cases arise here.

Case I When R = 150°,

`\implies` P + Q = 30°

`\implies` 0 < P, Q < 30°

`\implies` `sin P < 1/2, cos Q < 1`

`\implies` `3 sin P + 4 cos Q < 3/2 + 4`

`\implies` `3 sin P + 4 cos Q < 11/2 < 6`

`\implies` 3 sin P + 4 cos Q = 6, which is not possible.

Case II When R = 30°,

Hence R = 30° is the only possibility.