Question

$\sin x=\frac{6\sin {30}^o-8\cos {60}^o+2\tan {45}^o}{2({\sin }^2{30}^o+{\cos }^2{60}^o)}$, then $x=$ how much?

• A. ${30}^o$
• B. ${45}^o$
• C. ${60}^o$
• D. ${90}^o$

Answer 1

Answer: (D)

${90}^o$

$\sin x=\frac{6\sin {30}^o-8\cos {60}^o+2\tan {45}^o}{2({\sin }^2{30}^o+{\sin }^2{60}^o)}$
$\Rightarrow \sin x=\frac{6\times \frac{1}{2}-8\times 5\frac{1}{2}+2}{2({\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2)}=\frac{3.4+2}{2\left(\frac{1}{2}\right)}=\frac{1}{1}$
$\sin x=1$
$\therefore$ $x={90}^o$.

Hence, option D is correct.