Question

In $△$ PQR , $\angle$ Q = ${90}^{\circ }$. If cot P = $\sqrt{3}$ then, find the value of $\frac{sinP+cosR}{tanR}$.

• A. $\sqrt{3}$
• B. $\frac{1}{\sqrt{6}}$
• C. $\frac{2}{\sqrt{3}}$
• D. $\frac{1}{\sqrt{3}}$

Answer 1

The correct option is D $\frac{1}{\sqrt{3}}$

Given:
$\angle Q={90}^{\circ }$ and $cotP=\sqrt{3}$
We know that, $cot{30}^{\circ }$ = $\sqrt{3}$
$\therefore P$ = ${30}^{\circ }$

$\therefore \angle P+\angle Q+\angle R={180}^{\circ }$
($\because$ Sum of all angles of a triangle is ${180}^{\circ }$)
$\Rightarrow$ ${30}^{\circ }$ + ${90}^{\circ }$ + $\angle R={180}^{\circ }$
$\Rightarrow \angle R={180}^{\circ }-{30}^{\circ }-{90}^{\circ }$
$\therefore \angle R={60}^{\circ }$

Now, $sinP=sin{30}^{\circ }=\frac{1}{2}$,
$cosR=cos{60}^{\circ }=\frac{1}{2}$ ,
$tanR=tan{60}^{\circ }=\sqrt{3}$

$\therefore$ $\frac{sinP+cosR}{tanR}=\frac{sin{30}^{\circ }+cos{60}^{\circ }}{tan{60}^{\circ }}$

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

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