Question

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

Answer 1

Hint:Square both the equations and add them. Simplify the equation using the trigonometric identity ${{\cos }^{2}}x+{{\sin }^{2}}x=1$. Calculate the value of the angle $P+Q$ which satisfies the given equation. Use the fact that the sum of all the angles of a triangle is ${{180}^{\circ }}$ to calculate the measure of angle R.

Complete step-by-step answer:
We know that in $\Delta PQR$, we have $3\sin P+4\cos Q=6$ and $4\sin Q+3\cos P=1$. We have to calculate the measure of angle R.

We know the algebraic identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.
Squaring the equation $3\sin P+4\cos Q=6$, we have ${{\left( 3\sin P+4\cos Q \right)}^{2}}={{6}^{2}}$.
Thus, we have $9{{\sin }^{2}}P+16{{\cos }^{2}}Q+24\sin P\cos Q=36.....\left( 1 \right)$.
Similarly, squaring the equation $3\cos P+4\sin Q=1$, we have ${{\left( 3\cos P+4\sin Q \right)}^{2}}={{1}^{2}}$.
Thus, we have $9{{\cos }^{2}}P+16{{\sin }^{2}}Q+24\cos P\sin Q=1.....\left( 2 \right)$.
Adding equation (1) and (2), we have $9{{\sin }^{2}}P+16{{\cos }^{2}}Q+24\sin P\cos Q+9{{\cos }^{2}}P+16{{\sin }^{2}}Q+24\cos P\sin Q=36+1$.
Simplifying the above equation, we have \[9\left( {{\sin }^{2}}P+{{\cos }^{2}}P \right)+16\left( {{\sin }^{2}}Q+{{\cos }^{2}}Q \right)+24\left( \sin P\cos Q+\cos P\sin Q \right)=37\].
We know the trigonometric identity ${{\cos }^{2}}x+{{\sin }^{2}}x=1$.
Thus, we can rewrite the above equation as \[9+16+24\left( \sin P\cos Q+\cos P\sin Q \right)=37\].
So, we have $24\left( \sin P\cos Q+\cos P\sin Q \right)=37-\left( 9+16 \right)$.
$\begin{align}
  & \Rightarrow 24\left( \sin P\cos Q+\cos P\sin Q \right)=12 \\
 & \Rightarrow \sin P\cos Q+\cos P\sin Q=\dfrac{12}{24}=\dfrac{1}{2} \\
\end{align}$
We know the trigonometric identity $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$.
Thus, we have $\sin P\cos Q+\cos P\sin Q=\sin \left( P+Q \right)=\dfrac{1}{2}$.
So, the possible values of $P+Q$ are $P+Q={{\sin }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{\pi }{6},\dfrac{5\pi }{6}$.
We observe that $P+Q=\dfrac{\pi }{6}$ doesn’t satisfy the equation $3\cos P+4\sin Q=1$.
Thus, we have $P+Q=\dfrac{5\pi }{6}$.
We know that the sum of all interior angles of a triangle is $\pi $. Thus, we have $P+Q+R=\pi $.
$\begin{align}
  & \Rightarrow \dfrac{5\pi }{6}+R=\pi \\
 & \Rightarrow R=\pi -\dfrac{5\pi }{6}=\dfrac{\pi }{6} \\
\end{align}$
Hence, the measure of angle R is $\dfrac{\pi }{6}$, which is option (b).

Note: One must keep in mind that there are multiple solutions to the equation $P+Q={{\sin }^{-1}}\left( \dfrac{1}{2} \right)$. However, we will consider only those solutions whose value is less than or equal to $\pi $ as the sum of all interior angles of a triangle is $\pi $.Students should remember trigonometric identities and formulas for solving these types of problems.