Question

In triangle OPQ right angled at P, OP= 7 cm and OQ-PQ=1 cm. Determine the value of sinQ and cos Q.

Answer 1

let
OQ= x cm then
PQ = 1+x

by Pythagoras theorem we get
${op}^{2} + {pq}^{2} = { oq}^{2}$
$49 + {x}^{2} \: = {(1 + x)}^{2}$
on solving it we get value of x which is

x= 24

so PQ = 24
OQ = 25

as we know all 3 sides we can calculate sin Q and cos Q which are

sin Q = (7/25)
cos Q = (24/25)

Answer 2

Solution

In ∆ OPQ we have

OQ²=OP²+PQ²

- (pq+1)²=op²+pq²

PQ²+2PQ+1=OP²+PQ²

2PQ+1=49

2PQ=49-1

2PQ=48

PQ=48/2

PQ=24

OQ-PQ=1cm

OQ=(PQ+1)cm=25

Now, sinQ=OP/OQ=7/25

and,cosQ=PQ/QO=24/25