Question

In ∆ PQR , right-angled at Q, PR + QR = 25cm and PQ = 5cm. Determine the value of sin P, cos P and tan P.​

Answer 1

Step-by-step explanation:

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

Answer 2

$✅❤️Answer❤️$

⇒PQ=5cm

⇒PR+QR=25cm

⇒PR=25−QR

Now, In △PQR

⇒(PR)²

=PQ²+QR²

⇒(25−QR)²

=5²+QR²

⇒625+QR²−50QR=25+QR²

⇒50QR=600

⇒QR=12cm

⇒PR=25−12=13cm

$∴sinP= \frac{qr}{pr} = \frac{12}{13} ,cosP= \frac{pq}{pr} = \frac{5}{13} ,tanP= \frac{qr}{pq} = \frac{12}{5} \\ Hence, the answers are sinP= \\ \frac{12}{13} ,cosP= \: \frac{5}{13},tanP= \frac{12}{5}$