Question

If $\alpha ,\beta$ are solutions of $6\cos \theta +8\sin \theta =9$, then $\sin (\alpha +\beta )=$

• A. $\frac{3}{5}$
• B. $\frac{4}{5}$
• C. $\frac{24}{25}$
• D. $\frac{12}{13}$

Answer 1

The correct option is C

$\frac{24}{25}$

Explanation for the correct option:

Step 1. Find the value of $\sin (\alpha +\beta )$:

Given, $6\cos \theta +8\sin \theta =9$

$\Rightarrow$$2(3\cos \theta +4\sin \theta )=9$

$\Rightarrow$ $3\cos \theta +4\sin \theta =\frac{9}{2}$

Divide the above equation by 5:

$\frac{3}{5}\cos \theta +\frac{4}{5}\sin \theta =\frac{9}{10}$

Step 2. Let $\sin x=\frac{3}{5}$ and $\cos x=\frac{4}{5}$.(Consider)

$\sin x\cos \theta +\cos x\sin \theta =\frac{9}{10}$

$\Rightarrow$ $\sin (x+\theta )=\frac{9}{10}$

$\alpha ,\beta$ are the solutions of the given equation.

$\Rightarrow$$\sin (x+\theta )=\sin (x+\alpha )$ or $\sin (x+\theta )=\sin (x+\beta )$

$\begin{array}{rcl}x+\theta & =& x+\alpha \\ & =& \pi –(x+\alpha )\\ & =& x+\beta \end{array}$

$\therefore \alpha +\beta =\pi –2x$

Step 3. Take “sin” on both sides, we get

$\begin{array}{rcl}\sin (\alpha +\beta )& =& \sin (\pi –2x)\\ & =& \sin 2x\\ & =& 2\sin x\cos x\\ & =& 2\left(\frac{3}{5}\right)\left(\frac{4}{5}\right)\\ & =& \frac{\mathbf{24}}{\mathbf{25}}\end{array}$

Hence, Option ‘C’ is Correct.