Question

1. In ∆OPQ, right angled at P, OP=7cm and OQ-PQ= 1cm, then the values of sin Q .
(a)7/25
(b)24/25
(c)1
(d) none of the these​

Answer 1

Answer:

sorry dear I don't know plz forgive me

Answer 2

Answer:

The answer to this question is $\frac{7}{25}$

Step-by-step explanation:

Since ∆OPQ is a right-angled triangle

By Pythagoras theorem,

(OQ)² = (PQ)²+(OP)²

or, (OQ)² - (PQ)² = (OP)²

or, (OQ-PQ)(OQ+PQ) = (OP)²

or , (OQ+PQ) = (OP)²                     ( OQ-PQ = 1cm)

or, OQ+PQ = 49

Solving OQ-PQ = 1cm and OQ+PQ  = 49cm

we have,OQ = 25cm and PQ = 24cm

Now,

$sin{Q} = \frac{OP}{OQ}$

$or, sin{Q} = \frac{7}{25}$