In a \[\Delta PQR,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \] If \[3{\mkern 1mu} \sin {\mkern 1mu} P + 4{\mkern 1mu} \cos {\mkern 1mu} Q = 6\] and \[4\sin Q + 3\cos P = 1,\] then the angle R is equal to :
Solutions · 1
Official solution
Given 3 \[\sin \,P + 4\cos Q = 6\] \[\,\,\,\,\,\,\,\,...\left( i \right)\]
\[4\sin Q + 3\cos P = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)\]
Squaring and adding (i) \& (ii) we get
\[9\,{\sin ^2}P + 16{\cos ^2}Q + 24\sin P\cos Q\]
\[\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 16\,{\sin ^2}Q + 9{\cos ^2}P + 24\sin Q\cos P\]
\[\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\] =36+1=37
\[ \Rightarrow 9\left( {{{\sin }^2}p + {{\cos }^2}P} \right) + 16\left( {{{\sin }^2}Q + {{\cos }^2}q} \right)\]
\[\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 24\left( {\sin P\cos Q + \cos P\sin Q} \right) = 37\]
\[ \Rightarrow 9 + 16 + 24\sin \left( {P + Q} \right) = 37\]
[ As \[{\sin ^2}\theta + {\cos ^2}\theta = 1\] and
\[\sin A\cos B + \cos A\sin B\] \[ = \sin \left( {A + B} \right)\] ]
\[ \Rightarrow \sin \left( {P + Q} \right) = {1 \over 2}\]
\[ \Rightarrow P + Q = {\pi \over 6}\] or \[{{5\pi } \over 6}\]
\[ \Rightarrow R = {{5\pi } \over 6}\] or \[{\pi \over 6}\]
(as \[P + Q + R = \pi \] )
If \[R = {{5\pi } \over 6}\] then \[0 < P,Q < {\pi \over 6}\]
\[ \Rightarrow \cos Q < 1\] and \[\sin P < {1 \over 2}\]
\[ \Rightarrow 3\sin P + 4\cos Q < {{11} \over 2}\] which is not true.
So \[R = {\pi \over 6}\]