Question

In a trianglePQR, right angled at Q, PR+QR=25cm and PQ=5cm, find the value of sinP, cosP and tanP.

Answer 1

Answer:

Given PR + QR = 25 , PQ = 5

PR be x.  and QR = 25 - x 

Pythagoras theorem ,PR2 = PQ2 + QR2

x2 = (5)2 + (25 - x)2

x2 = 25 + 625 + x2 - 50x

50x = 650

x = 13

 

 PR = 13 cm

QR = (25 - 13) cm = 12 cm

sin P = QR/PR = 12/13

cos P = PQ/PR = 5/13

tan P = QR/PQ = 12/5 

Hope it helps you

Answer 2

SOLUTION

It is given that

PR+QR= 25cm

PR= 25-QR..........(1)

by Pythagoras Theorem

=)PR^2 =QR^2 +PQ^2

=)(25-QR)^2 = QR^2 +5^2

=)625 +QR^2-50 QR = QR^2 +25

Solving for QR,we get

=) QR= 12cm

Then from (1)

PR= 25-12 = 13cm

Now from ∆RQP

sinP= QR/PR= 12/13

cosP= PQ/PR= 5/13

tanP= QR/PQ= 12/5

Hope it helps ✔️