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In $ΔPQR$, right-angled at $Q, PR + QR = 25\ cm$ and $PQ = 5\ cm$. Determine the values of $sin\ P, cos\ P$ and $tan\ P$.
Given:
In $ΔPQR$, right-angled at $Q, PR + QR = 25\ cm$ and $PQ = 5\ cm$.
To do:
We have to find the values of $sin\ P, cos\ P$ and $tan\ P$.
Solution:
We know that,
In a right-angled triangle $PQR$ with a right angle at $Q$,
By Pythagoras theorem,
$PR^2=PQ^2+QR^2$
$\mathrm{PQ}^{2}=\mathrm{PR}^{2}-\mathrm{QR}^{2}$
$(5)^{2} =(\mathrm{PR}+\mathrm{QR})(\mathrm{PR}-\mathrm{QR})$
$25=25(\mathrm{PR}-\mathrm{QR})$
$\mathrm{PR}-\mathrm{QR}=1$
$\mathrm{PR}+\mathrm{QR}=25$
This implies,
$\mathrm{PR}-\mathrm{QR}+\mathrm{PR}+\mathrm{QR}=1+25$
$=26$
$2PR=26$
$PR=13$
$\Rightarrow QR=PR-1$
$=13-1$
$=12$
By trigonometric ratios definitions,
$sin\ P=\frac{Opposite}{Hypotenuse}=\frac{QR}{PR}$
$=\frac{12}{13}$
$cos\ P=\frac{Adjacent}{Hypotenuse}=\frac{PQ}{PR}$
$=\frac{5}{13}$
$tan\ P=\frac{Opposite}{Adjacent}=\frac{QR}{PQ}$
$=\frac{12}{5}$