Question

In $∆PQR$, right-angled at $Q$, $PR+QR=25cm$ and $PQ=5cm$. Determine the values of$\sin P,\cos P$ and $\tan P$?

Answer 1

Step 1: Given data

• $PR+QR=25cm.....\left(i\right)$
• $PQ=5cm$

Step 2: Apply Pythagoras theorem

$$\begin{gathered} P{Q}^2+Q{R}^2=P{R}^2 \\ \Rightarrow 25=P{R}^2-Q{R}^2 \\ \Rightarrow 25=(PR-QR)(PR+QR) \end{gathered}$$

from equation $\left(i\right)$

$\Rightarrow PR-QR=1\dots \dots \left(ii\right)\left[{\mathit{a}}^{\mathbf{2}}\mathbf{-}{\mathit{b}}^{\mathbf{2}}\mathbf{=}\mathbf{(}\mathit{a}\mathbf{-}\mathit{b}\mathbf{)}\mathbf{(}\mathit{a}\mathbf{+}\mathit{b}\mathbf{)}\right]$

Step 3: Determine $\sin P,\cos P$ and $\tan P$

Adding $\left(i\right)$ and $\left(ii\right)$ we get,

$\begin{array}{rcl}\Rightarrow (PR+QR)+(PR-QR)& =& 25+1\\ \Rightarrow 2PR& =& 26\\ \Rightarrow PR& =& 13\\ \Rightarrow QR& =& 12\mathbf{[}\mathit{F}\mathit{r}\mathit{o}\mathit{m}\mathbf{}\mathbf{(}\mathit{i}\mathit{i}\mathbf{\left)}\mathbf{\right]}\\ & & \end{array}$

$$\begin{gathered} \Rightarrow \sin \left(P\right)=\frac{QR}{PR}=\frac{12}{13}\mathbf{[}\mathbf{s}\mathbf{i}\mathbf{n}\left(\theta \right)\mathbf{=}\frac{\mathbf{Perpendicular}}{\mathbf{H}\mathbf{y}\mathbf{p}\mathbf{o}\mathbf{t}\mathbf{e}\mathbf{n}\mathbf{u}\mathbf{s}\mathbf{e}}\mathbf{]} \\ \Rightarrow \cos \left(P\right)=\frac{PQ}{PR}=\frac{5}{13}\mathbf{[}\mathbf{c}\mathbf{o}\mathbf{s}\left(\theta \right)\mathbf{=}\frac{\mathbf{Base}}{\mathbf{H}\mathbf{y}\mathbf{p}\mathbf{o}\mathbf{t}\mathbf{e}\mathbf{n}\mathbf{u}\mathbf{s}\mathbf{e}}\mathbf{]} \\ \Rightarrow \tan \left(P\right)=\frac{\sin \left(P\right)}{\cos \left(P\right)}=\frac{12}{5}\mathbf{}\mathbf{[}\mathbf{t}\mathbf{a}\mathbf{n}\left(\theta \right)\mathbf{=}\frac{\mathbf{Perpendicular}}{\mathbf{Base}}\mathbf{]} \end{gathered}$$

Therefore, $\sin P=\frac{12}{13},\cos P=\frac{5}{13}$ and $\tan P=\frac{12}{5}$