In ∆ PQR , right-angled at Q, PR + QR = 25cm and PQ = 5cm. Determine the value of sin P, cos P and tan P. About the author Ayla
[tex]✅❤️Answer❤️[/tex] ⇒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 [tex]∴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} [/tex] Reply
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 Reply
[tex]✅❤️Answer❤️[/tex]
⇒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
[tex]∴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} [/tex]
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