Question

In a triangle PQR. right-angled at Q. PR +QR= 25 cm PQ=5 cm. then the value of sin P is
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Answer 1

Answer:

PQ=5cm

⇒PR+QR=25cm

⇒PR=25−QR

Now, In △PQR

⇒(PR)

2

=PQ

2

+QR

2

⇒(25−QR)

2

=5

2

+QR

2

⇒625+QR

2

−50QR=25+QR

2

⇒50QR=600

⇒QR=12cm

⇒PR=25−12=13cm

∴sinP=

PR

QR

=

13

12

,cosP=

PR

PQ

=

13

5

,tanP=

PQ

QR

=

5

12

Hence, the answers are sinP=

13

12

,cosP=

13

5

,tanP=

5

12

.