Question

In a right angled triangle ABC, right angled at C, $\tan B$ $=$ $2\sin B$ , 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

Answer: (A)

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

$\Rightarrow Given:\tan B=2\sin B$

$\Rightarrow \frac{\sin B}{\cos B}=2\sin B$

$\Rightarrow \cos B=\frac{1}{2}$

$\Rightarrow B={60}^0$

Therefore $A={90}^0-{60}^0={30}^0$

$\Rightarrow \sin A+\sin B$

$=\sin {30}^0+\sin {60}^0$

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

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

Hence answer is $A$