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 ______.

विकल्प

• ${\displaystyle \frac{5\pi }{6}}$

• ${\displaystyle \frac{\pi }{6}}$

• ${\displaystyle \frac{\pi }{4}}$

• ${\displaystyle \frac{3\pi }{4}}$

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 ${\displaystyle \underline{\frac{\pi }{6}}}$.

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

${\displaystyle \Rightarrow }$ 24 [sin (P + Q)] = 37 – 25

${\displaystyle \Rightarrow }$ sin (P + Q) = ${\displaystyle \frac{1}{2}}$

Since, P and Q are angles of ΔPQR.

0° < P, Q < 180°

${\displaystyle \Rightarrow }$ P + Q = 30° or 150°

${\displaystyle \Rightarrow }$ R = 150° or 30°   ...[respectively]

Hence, two cases arise here.

Case I When R = 150°,

${\displaystyle \Rightarrow }$ P + Q = 30°

${\displaystyle \Rightarrow }$ 0 < P, Q < 30°

${\displaystyle \Rightarrow }$ ${\displaystyle \sin P<\frac{1}{2},\cos Q<1}$

${\displaystyle \Rightarrow }$ ${\displaystyle 3\sin P+4\cos Q<\frac{3}{2}+4}$

${\displaystyle \Rightarrow }$ ${\displaystyle 3\sin P+4\cos Q<\frac{11}{2}<6}$

${\displaystyle \Rightarrow }$ 3 sin P + 4 cos Q = 6, which is not possible.

Case II When R = 30°,

Hence R = 30° is the only possibility.