Question

In ∆PQR, right angled at Q, PQ = 4 cm and RQ = 3 cm. Find the values of sin P, sin R, sec P and sec R.

Answer 1

SOLUTION :  

Given:

ΔPQR is right angled at vertex Q.

PQ= 4cm

RQ= 3cm

In ΔPQR,

PR² = PQ² + RQ²

[by using Pythagoras theorem]

PR² = 4² +3²

PR² = 16 + 9

PR² = 25

PR = √25

PR = 5

Hence, Hypotenuse = 5

With reference to ∠P :  

sin P =  Perpendicular / Hypotenuse

sin P = RQ / PR

sin P =3/5

With reference to ∠R :  

sin R =  Perpendicular / Hypotenuse

sin R = PQ/PR

sin R = 4/5

With reference to ∠P :

sec P = Hypotenuse/ Base

sec P = PR/PQ

sec P = 5/4

With reference to ∠R :

sec R= Hypotenuse/Base

sec R = PR/RQ

sec R = 5/3

Hence, sin P= 3/5 , sin R = 4/5,

sec P = 5/4, sec R = 5/3

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

By Pythagoras theorum
h^2=p^2+b^2
h^2=3^2+4^2
h^2=9+16
h^2=25
h=5

sin R=p/h=4/5
sec P=h/b=5/4
sec R=5/3