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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 t...

In a ΔPQR\Delta PQR, if 3sin⁡P+4cos⁡Q=63\sin P+4\cos Q=6 and 4sin⁡Q+3cos⁡P=14\sin Q+3\cos P=1, then the angle R is equal to
(a) 5π6\dfrac{5\pi }{6}
(b) π6\dfrac{\pi }{6}
(c) π4\dfrac{\pi }{4}
(d) 3π4\dfrac{3\pi }{4}

Explanation

Solution

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

Complete step-by-step answer:
We know that in ΔPQR\Delta PQR, we have 3sin⁡P+4cos⁡Q=63\sin P+4\cos Q=6 and 4sin⁡Q+3cos⁡P=14\sin Q+3\cos P=1. We have to calculate the measure of angle R.

We know the algebraic identity (a+b)2=a2+b2+2ab{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab.
Squaring the equation 3sin⁡P+4cos⁡Q=63\sin P+4\cos Q=6, we have (3sin⁡P+4cos⁡Q)2=62{{\left( 3\sin P+4\cos Q \right)}^{2}}={{6}^{2}}.
Thus, we have 9sin⁡2P+16cos⁡2Q+24sin⁡Pcos⁡Q=36.....(1)9{{\sin }^{2}}P+16{{\cos }^{2}}Q+24\sin P\cos Q=36.....\left( 1 \right).
Similarly, squaring the equation 3cos⁡P+4sin⁡Q=13\cos P+4\sin Q=1, we have (3cos⁡P+4sin⁡Q)2=12{{\left( 3\cos P+4\sin Q \right)}^{2}}={{1}^{2}}.
Thus, we have 9cos⁡2P+16sin⁡2Q+24cos⁡Psin⁡Q=1.....(2)9{{\cos }^{2}}P+16{{\sin }^{2}}Q+24\cos P\sin Q=1.....\left( 2 \right).
Adding equation (1) and (2), we have 9sin⁡2P+16cos⁡2Q+24sin⁡Pcos⁡Q+9cos⁡2P+16sin⁡2Q+24cos⁡Psin⁡Q=36+19{{\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(sin⁡2P+cos⁡2P)+16(sin⁡2Q+cos⁡2Q)+24(sin⁡Pcos⁡Q+cos⁡Psin⁡Q)=379\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⁡2x+sin⁡2x=1{{\cos }^{2}}x+{{\sin }^{2}}x=1.
Thus, we can rewrite the above equation as 9+16+24(sin⁡Pcos⁡Q+cos⁡Psin⁡Q)=379+16+24\left( \sin P\cos Q+\cos P\sin Q \right)=37.
So, we have 24(sin⁡Pcos⁡Q+cos⁡Psin⁡Q)=37−(9+16)24\left( \sin P\cos Q+\cos P\sin Q \right)=37-\left( 9+16 \right).
⇒24(sin⁡Pcos⁡Q+cos⁡Psin⁡Q)=12 ⇒sin⁡Pcos⁡Q+cos⁡Psin⁡Q=1224=12 \begin{aligned} & \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{aligned}
We know the trigonometric identity sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B.
Thus, we have sin⁡Pcos⁡Q+cos⁡Psin⁡Q=sin⁡(P+Q)=12\sin P\cos Q+\cos P\sin Q=\sin \left( P+Q \right)=\dfrac{1}{2}.
So, the possible values of P+QP+Q are P+Q=sin⁡−1(12)=π6,5π6P+Q={{\sin }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{\pi }{6},\dfrac{5\pi }{6}.
We observe that P+Q=π6P+Q=\dfrac{\pi }{6} doesn’t satisfy the equation 3cos⁡P+4sin⁡Q=13\cos P+4\sin Q=1.
Thus, we have P+Q=5π6P+Q=\dfrac{5\pi }{6}.
We know that the sum of all interior angles of a triangle is π\pi . Thus, we have P+Q+R=πP+Q+R=\pi .
⇒5π6+R=π ⇒R=π−5π6=π6 \begin{aligned} & \Rightarrow \dfrac{5\pi }{6}+R=\pi \\\ & \Rightarrow R=\pi -\dfrac{5\pi }{6}=\dfrac{\pi }{6} \\\ \end{aligned}
Hence, the measure of angle R is π6\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(12)P+Q={{\sin }^{-1}}\left( \dfrac{1}{2} \right). However, we will consider only those solutions whose value is less than or equal to π\pi as the sum of all interior angles of a triangle is π\pi .Students should remember trigonometric identities and formulas for solving these types of problems.