Question

In the given figure, Find $RP$ and $PS$ using the information given in $△PSR$

Answer 1

Given $\angle P={30}^{\circ }$
Since, we have $sinP=\frac{SR}{PR}$
$\Rightarrow sin\left(30\right)=\frac{6}{PR}$
$\Rightarrow \frac{1}{2}=\frac{6}{PR}$

$\Rightarrow PR=6\times 2$
$\therefore PR=12$ units.
Similar way, consider,$cos\left(30\right)=\frac{SP}{PR}$
$\Rightarrow cos\left(30\right)=\frac{SP}{12}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{SP}{12}$
$\Rightarrow SP=\frac{12\times \sqrt{3}}{2}$
$\therefore SP=6\sqrt{3}$ units.

Alternatively,

In $△PSR$,

$\angle S={90}^{\circ },\angle P={30}^{\circ },$

Since, we have $\angle S+\angle P+\angle R={180}^{\circ }$
$\Rightarrow 90+30+\angle R=180$
$\Rightarrow angleR=180-110$

$\therefore \angle R={60}^{\circ }$

By ${30}^{\circ }-{60}^{\circ }-{90}^{\circ }$therom, $RS=\frac{1}{2}\times RP$

$\Rightarrow 6=\frac{1}{2}\times RP$

$\Rightarrow 6\times 2=RP$

$\Rightarrow RP=\mathrm{12...}\left(1\right)$

$PS=\frac{\sqrt{3}}{2}\times RP$

$=\frac{\sqrt{3}}{2}\times 12$

$=6\sqrt{3}...\left(2\right)$

Hence, $RP=12$ and $PS=6\sqrt{3}$