Question

In a triangle PQR, if 3 sin P + 4 cos Q = 6, and 4 sin Q + 3 cos P = 1, then the angle R is equal to:
$\left( a \right)\dfrac{{5\pi }}{6}$
$\left( b \right)\dfrac{\pi }{6}$
$\left( c \right)\dfrac{\pi }{4}$
$\left( d \right)\dfrac{{3\pi }}{4}$

Answer 1

Hint: In this particular question use the concept of squaring the equation on both sides and them, later on in the solution use the concept that in any triangle the sum of all angles should be equal to 180 degrees so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given data
In a triangle PQR,
3 sin P + 4 cos Q = 6................ (1),
And 4 sin Q + 3 cos P = 1.................... (2)
Now squaring on both sides of the equations and add them we have,
$ \Rightarrow {\left( {3\sin P + 4\cos Q} \right)^2} + {\left( {4\sin Q + 3\cos P} \right)^2} = {6^2} + {1^2}$
Now expand the square according to the property that ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ so we have,
$ \Rightarrow \left( {9{{\sin }^2}P + 16{{\cos }^2}Q + 24\sin P\cos Q} \right) + \left( {16{{\sin }^2}Q + 9{{\cos }^2}P + 24\sin Q\cos P} \right) = 36 + 1$
Now simplify this we have,
\[ \Rightarrow 9\left( {{{\sin }^2}P + {{\cos }^2}P} \right) + 16\left( {{{\sin }^2}Q + {{\cos }^2}Q} \right) + 24\left( {\sin P\cos Q + \sin Q\cos P} \right) = 37\]
Now as we know that ${\sin ^2}x + {\cos ^2}x = 1,\left( {\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B} \right)$, so use these properties in the above equation we have,
\[ \Rightarrow 9\left( 1 \right) + 16\left( 1 \right) + 24\sin \left( {P + Q} \right) = 37\]
\[ \Rightarrow 24\sin \left( {P + Q} \right) = 37 - 25 = 12\]
\[ \Rightarrow \sin \left( {P + Q} \right) = \dfrac{{12}}{{24}} = \dfrac{1}{2}\].................... (1)
Now as we know that in any triangle the sum of all angles is equal to 180 degrees so we have,
$ \Rightarrow P + Q + R = {180^o}$
$ \Rightarrow P + Q = {180^o} - R$
Now substitute this value in equation (1) we have,
\[ \Rightarrow \sin \left( {{{180}^o} - R} \right) = \dfrac{1}{2}\]
Now take sin to LHS we have,
\[ \Rightarrow {180^o} - R = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right)\]
\[ \Rightarrow {180^o} - R = {30^o}\],\[{180^o} - R = {150^o}\] $\left[ {\because {{\sin }^{ - 1}}\left( {\dfrac{1}{2}} \right) = {{30}^o},{{150}^o}} \right]$
$ \Rightarrow R = {150^o},{30^o}$
Now convert into radian so multiply by $\left( {\dfrac{\pi }{{{{180}^o}}}} \right)$ we have,
$ \Rightarrow R = {150^o} \times \dfrac{\pi }{{{{180}^o}}} = \dfrac{{5\pi }}{6}$, $R = {30^o} \times \dfrac{\pi }{{{{180}^o}}} = \dfrac{\pi }{6}$.
So this is the required value of the angle R.
Hence options (a) and (b) are the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the basic trigonometric identities such as ${\sin ^2}x + {\cos ^2}x = 1,\left( {\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B} \right)$ and the value of standard sine angle and always recall that to convert the degree into radian we have to multiply $\left( {\dfrac{\pi }{{{{180}^o}}}} \right)$.