Question

Example 5 : In A OPQ, right-angled at P.
OP = 7 cm and OQ - PQ = 1 cm (see Fig. 8.12).
Determine the values of sin Q and cos Q.​

Answer 1

Answer:

(x-1)/x,7/x

Step-by-step explanation:

sinQ=p/h=(x-1)/x, cosQ=b/h= 7/x

Answer 2

Answer:

Sin(Q) = 7/25 & Cos(q) = 24/25

Step-by-step explanation:

Solution:-

In given that in triangle OPQ

OQ - PQ = 1cm

OQ = 1 + PQ

using Pythagoras theorem,

(Hypotenuse)² = (height)² + ( base)²

OQ ² = PQ² + OP²

=> (1+PQ)² = PQ² + 7²

=> 1 + PQ²+2PQ = PQ² + 49

=> 1+ PQ²+2PQ-PQ²-49 = 0

=> 2PQ- 48 =0

=> 2PQ = 48

$= > pq \: = \frac{48}{2}$

=> PQ = 24 cm

Also We find,

OQ = 1 + PQ

=> OQ = 1 + 24

=> OQ = 25 cm

Now,

$\sin(q) = \frac{op}{pq}$

$= > \sin(q) = \frac{7}{25}$

Again,

$\cos(q) = \frac{pq}{oq}$

$= > \cos(q) = \frac{24}{25}$

Hance,

$\sin(q ) = \frac{7}{25}$

and,

$\cos(q) = \frac{24}{25}$