Question

In a ΔPQR, if 3sinP + 4cosQ = 6 and 4sinQ + 3cosP = 1, then the angle R is equal to:

• A. $\frac{5π}{6}$
• B. $\frac{π}{6}$
• C. $\frac{π}{4}$
• D. $\frac{3π}{4}$

Answer 1

Answer: (B)

$\frac{π}{6}$

Answer 2

$\text{Given a}\,\triangle \text{PQR such that}$

3sinP + 4cosQ = 6 ….(i)
4sinQ + 3cosP = 1 .…(ii)

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

$(3sinP + 4cosQ)^2 + (4sinQ + 3cosP)^2\, =36+1$
⇒ $9(sin^2P + cos^2P) + 16(sin^2Q + cos^2Q) + 2\times3 \times4 (sinP cosQ + sinQcosP) = 37$
⇒ $24[sin(P+Q)] = 37-25$
⇒ $sin(P+Q) = \frac{1}{2}$

Since, P and Q are angles of $\triangle PQR$, hence 0^\degree<P,\, Q<180^\degree
⇒ P+Q = 30^\degree\,\, or \,\,150^\degree
⇒ R = 150^\degree \,\, or \,\, 30^\degree

Hence, Two cases aries here,
**Case I **R = 150^\degree
⇒P+Q = 30^\degree
⇒ 0< P,\, Q<30^\degree
⇒ $sinP<\frac{1}{2}, cosQ<1$
⇒ $3sinP + 4cosQ <\frac{3}{2} + 4$
⇒ $3sinP + 4cosQ <\frac{11}{2} < 6$
⇒ $3sinP + 4cosQ ⇒6\,\, \text{is not possible.}$

**Case II **R = 30^\degree
\text{Hence, R} = 30^\degree \text{is the only possibility. }

So, The correct option is (B) $\frac{π}{6}$.