Question

Ques:- In triangle PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.

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Answer 1

⇒PQ=5cm

⇒PR+QR=25cm

⇒PR=25−QR

Now, In △PQR

$⇒(PR) {}^{2} =PQ {}^{2} +QR {}^{2}$

$⇒(25−QR) {}^{2} =52+QR {}^{2} \\ </p><p></p><p>⇒625+qr {}^{2} - 50QR \\ =25+QR2 \\ </p><p></p><p>⇒50QR=600 </p><p></p><p>$

$QR=12cm \\ </p><p></p><p>⇒PR=25−12=13cm</p><p></p><p>$

$sin \: p = \frac{qr}{pr} = \frac{12}{13} ,cos \: p = \frac{pq}{pr} = \\ \frac{5}{13} ,tan \: p = \frac{qr}{pq} = \frac{12}{5} \\ </p><p>Hence, the answers are sinP=$

$\frac{12}{13},cos \: p = \frac{5}{13} ,tan \: p = \frac{12}{5}$

Answer 2

⇒PQ=5cm

⇒PR+QR=25cm

⇒PR=25−QR

Now, In △PQR

⇒(PR)

2=PQ

2 +QR

⇒(25−QR)

2=52+QR

⇒625+QR

⇒50QR=600

⇒QR=12cm

⇒PR=25−12=13cm

HENCE AN

13

12

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