Question

In $△OPQ$, right-angled at $P,OP=7cm$ and $OQ-PQ=1cm$, determine the values of $\sin Q$ and $\cos Q$.

Answer 1

Given

$OP=7cm$

$OQ-PQ=1cm$

$\begin{array}{c}In\Delta OPQ\\ O{Q}^2=O{P}^2+P{Q}^2\left(ByPythagorusTheorem\right)\\ \Rightarrow O{Q}^2=7^2+P{Q}^2\\ \Rightarrow {\left(1+PQ\right)}^2=49+P{Q}^2\left(From\left(1\right)\right)\\ \Rightarrow 1+{\left(PQ\right)}^2+2PQ=49+P{Q}^2\\ \Rightarrow 2PQ=48\\ \Rightarrow PQ=24cm\end{array}$

Now

$\begin{array}{c}OQ=1+PQ\\ =1+24\\ =25cm\end{array}$

So,

$\sin Q=\frac{OP}{OQ}=\frac{7}{25}$
$\cos Q=\frac{PQ}{OQ}=\frac{24}{25}$