Question

If the angles $A\text{and}B$ of the triangle $ABC$ satisfy the equation $\sin A+\sin B=\sqrt{3}\left(\cos B-\cos A\right)$, then they differ by

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is B

$\frac{\pi }{3}$

Explanation for the correct option

Simplify the given equation

$\begin{array}{rcl}\therefore \sin A+\sin B& =& \sqrt{3}\cos B-\sqrt{3}\cos A\\ & \Rightarrow & \sin A+\sqrt{3}\cos A=\sqrt{3}\cos B-\sin B\end{array}$

Divide both the sides by $2$ to get

$\frac{1}{2}\sin \left(A\right)+\frac{\sqrt{3}}{2}\cos \left(A\right)=\frac{\sqrt{3}}{2}\cos \left(B\right)-\frac{1}{2}\sin \left(B\right)$

We know that $\sin \frac{\pi }{6}=\frac{1}{2}\text{and}\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}$, then we have

$\sin \left(\frac{{\displaystyle \pi }}{{\displaystyle 6}}\right)\sin \left(A\right)+\cos \left(\frac{\pi }{6}\right)\cos \left(A\right)=\cos \left(\frac{\pi }{6}\right)\cos \left(B\right)-\sin \left(\frac{\pi }{6}\right)\sin \left(B\right)$

Now, apply the identities $\mathbf{cos}\left(\theta -\alpha \right)\mathbf{=}\mathbf{cos}\mathit{\theta }\mathbf{cos}\mathit{\alpha }\mathbf{+}\mathbf{sin}\mathit{\theta }\mathbf{sin}\mathit{\alpha }$ and $\mathbf{cos}\left(\theta +\alpha \right)\mathbf{=}\mathbf{cos}\mathit{\theta }\mathbf{cos}\mathit{\alpha }\mathbf{-}\mathbf{sin}\mathit{\theta }\mathbf{sin}\mathit{\alpha }$

$\begin{array}{rcl}\therefore \cos \left(A-\frac{\pi }{6}\right)& =& \cos \left(\frac{\pi }{6}+B\right)\\ & \Rightarrow & A-\frac{\pi }{6}=\frac{\pi }{6}+B\\ & \Rightarrow & A-B=\frac{\pi }{6}+\frac{\pi }{6}\\ & \Rightarrow & A-B=\frac{\pi }{3}\end{array}$

Hence, the correct option is (B).