Question

In \(\triangle ABC\), if \[ 3\sin A+4\cos B=6 \] and \[ 4\sin B+3\cos A=1, \] then the angle \(C\) is

• A. \(\dfrac{\pi}{2}\)
• B. \(\dfrac{\pi}{3}\)
• C. \(\dfrac{\pi}{4}\)
• D. \(\dfrac{\pi}{6}\)

Hint:

In triangle trigonometry problems, always use \(A+B+C=\pi\). This relation helps connect separate equations involving \(A\), \(B\), and \(C\).

Answer 1

The Correct Option is D

Step 1: Square and add both equations.
\((3\sin A+4\cos B)^2+(4\sin B+3\cos A)^2=36+1=37\). Expanding: \(9\sin^2A+24\sin A\cos B+16\cos^2B+16\sin^2B+24\sin B\cos A+9\cos^2A=37\). Simplify: \(9+16+24(\sin A\cos B+\cos A\sin B)=37\Rightarrow 24\sin(A+B)=12\Rightarrow\sin(A+B)=\frac{1}{2}\).

Step 2: Find angle C using A + B + C = pi.
\(\sin(A+B)=\sin(\pi-C)=\sin C=\frac{1}{2}\Rightarrow C=\dfrac{\pi}{6}\). \[ \boxed{\dfrac{\pi}{6}} \]