Question

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

• 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 cot P = $\sqrt{3}$
$\because$ cot ${30}^{\circ }$ = $\sqrt{3}$
$\therefore$ P = ${30}^{\circ }$

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

Now, sin P = sin ${30}^{\circ }$ = $\frac{1}{2}$,
cos R = cos ${60}^{\circ }$ = $\frac{1}{2}$ ,
tan R = tan ${60}^{\circ }$ = $\sqrt{3}$

$\therefore$ $\frac{sinP+cosR}{tanR}$

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

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