Question

In a right angled triangle ABC right angled at C, $\tan B=\sqrt{3}$ then $\sin A+\sin B$ is equal to

• A. $\frac{\sqrt{3}+1}{2}$
• B. $\sqrt{3}$
• C. {$\frac{1}{2}$}
• D. 1

Answer 1

The correct option is A $\frac{\sqrt{3}+1}{2}$

In the triangle $ABC$,

$\tan B=\sqrt{3}$

$\sin A=\frac{1}{2}$

$\sin B=\frac{\sqrt{3}}{2}$

Since,

$=\sin A+sinB$

$=\frac{1}{2}+\frac{\sqrt{3}}{2}$

$=\frac{\sqrt{3}+1}{2}$

Hence, this is the answer.