Question

In $△PQR$ right angled at $Q$, $PR+QR=25cm$ and $PQ=5cm$ determine the values of $\sin \left(P\right)$.

Answer 1

Step 1: Calculate the length of $PR$ and $QR$.

It is given that $PR+QR=25cm$ and $PQ=5cm$.

Using Pythagoras Theorem in right angled $△PQR$.

$\Rightarrow P{R}^2=P{Q}^2+Q{R}^2$

Let $PR=xcm$ and $QR=\left(25-x\right)cm$.

$\begin{array}{rcl}& \Rightarrow & x^2=5^2+{\left(25-x\right)}^2\\ & \Rightarrow & x^2=25+625+x^2-50x\\ & \Rightarrow & 50x=650\\ & \Rightarrow & x=13cm\end{array}$

So, $PR=13cm$and

$\begin{array}{rcl}QR& =& 25-13cm\\ & =& 12cm\end{array}$

Step 2: Calculate the value of $\sin \left(P\right)$

$\begin{array}{rcl}\sin \left(P\right)& =& \frac{perp}{hyp}\\ & =& \frac{QR}{PR}\\ & =& \frac{12}{13}\end{array}$

Hence, the value of $\sin P=\frac{12}{13}$.