Question

Solve $6\sin \theta +7\cos \theta =9$ if

• A. $\sin \theta =\frac{4}{3}$
• B. $\sin \theta =\frac{7}{8}$
• C. $\sin \theta =\frac{8}{15}$
• D. $\sin \theta =\frac{8}{17}$

Answer 1

The correct option is C $\sin \theta =\frac{8}{15}$

given, $6\left(\frac{2tan\frac{x}{2}}{1+ta{n}^2\frac{x}{2}}\right)+7\left(\frac{1-ta{n}^2\frac{x}{2}}{1+ta{n}^2\frac{x}{2}}\right)=9$

$\Rightarrow 12tan\frac{x}{2}+7=9+16ta{n}^2\frac{x}{2}$

$\Rightarrow 16ta{n}^2\frac{x}{2}-12tan\frac{x}{2}+2=0$

$\Rightarrow tan\frac{x}{2}=\frac{1}{2};tan\frac{x}{2}=\frac{1}{4}$

$weknowsinx=\frac{2tan\frac{x}{2}}{1-ta{n}^2\frac{x}{2}}$

$hencesinx=\frac{4}{3}\left(notpossible\right);sinx=\frac{8}{15}$

$herex=\theta$

So, $\sin \theta =8/15$