Question

In the given figure $△PQR$, right-angled at $Q,QR=9cm$ and $PR-PQ=1cm$. Determine the value of $\sin R+\cos R$.

Answer 1

R.E.F. Image.

Given $\Delta PQR$ is right angle triangle with $QR=9cm$

& $PR-PQ=1cm$

then $PR=(1+PQ)$

squaring both sides

$(PR{)}^2=(1+PQ{)}^2=1+(PQ{)}^2+2PQ$

$(PR{)}^2-(PQ{)}^2=1+2PQ$

$(QR{)}^2=1+2PQ$ $[P{R}^2=P{Q}^2+Q{R}^2]$

$81-1=2PQ$

$[PQ=40cm]$

now, using Pythagoras theorem. or equation (1) we have

$PR-PQ=1$

$[PR=41cm]$

now $sinR=\frac{PQ}{PR}=\frac{40}{41}cosR=\frac{9}{41}=\frac{QR}{PR}$

then $sinR+cosR=\frac{40}{41}+\frac{9}{41}=\frac{49}{41}.$

$sinR+cosR=\frac{49}{41}.$