Question

In triangle PQR right angled at Q , PR+QR=25cm determine the values of sin P , cos P , tan P.

Answer 1

Here's the answer.....but ur question is incomplete...... It's there in NCERT book.

Answer 2

$\rm { Given \: that, }$

$\rm { PR + QR = 25}$

$\rm { PQ = 5 }$

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$\rm { Let \: PR \: be \: x.</p><p>}$

$\rm { \therefore QR = 25 - x</p><p> }$

Applying Pythagoras theorem in ΔPQR, we obtain,

$\rm\blue { {PR}^{2} = {PQ}^{2} + {QR}^{2} }$

$\rm{ \leadsto {x}^{2} = {(5)}^{2} + {(25-x)}^{2} }$

$\rm{ \leadsto \cancel { {x}^{2}} = 25 + 625 \cancel { + {x}^{2}} - 50x }$

$\rm{ \leadsto 50x = 650</p><p> }$

$\rm{ \leadsto x = \cancel {\dfrac{650}{50}</p><p>}=13cm }$

$\rm\red { \therefore, PR = 13 cm</p><p>}$

$\rm\red { QR = (25 - 13) cm = 12 cm }$

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Now, the values of sin P, cos P and tan P

$\rm { sin \: P =\dfrac{ side \: opposite \: to \: \angle P}{Hypotenuse} = \dfrac{12}{13} }$

$\rm { cos \: P = \dfrac{ side \: adjacent \: to \: \angle P}{Hypotenuse} = \dfrac{5}{13} }$

$\rm { tan \: P= \dfrac{ side \: opposite \: to \: \angle P}{side \: adjacent \: to \: \angle P } = \dfrac{12}{5} }$