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

The Correct Option is B

\(\text{Given a}\,\triangle \text{PQR such that}\)

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) \(\frac{π}{6}\).