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

AcademicMathematicsNCERTClass 10

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,

$sin\ P=\frac{Opposite}{Hypotenuse}=\frac{QR}{PR}$

$=\frac{12}{13}$

$cos\ P=\frac{Adjacent}{Hypotenuse}=\frac{PQ}{PR}$

$=\frac{5}{13}$

$tan\ P=\frac{Opposite}{Adjacent}=\frac{QR}{PQ}$

$=\frac{12}{5}$

raja
Updated on 10-Oct-2022 13:22:12

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