Question

In ∆ OPQ, right angled at P, OP = 7cm and oq= pq =1cm determine the value of sin q and cos q​

Answer 1

Correct question:

In ∆ OPQ, right angled at P, OP = 7cm and OQ-PQ = 1cm. Determine the values of sin Q and cos Q.

$\huge{ \underline{ \mathbb{SOLUTION:-}}}$

Given:

• In ∆ OPQ, right angled at P, OP = 7cm and OP = 7cm and OQ - PQ = 1cm.

Need to do:

• Determining the values of sin Q and cos Q.

Answer:

• sin Q = 7/25 and cos Q = 24/25.

Explanation:

In ∆ OPQ, we have

$\small{ \implies{ {OP}^{2} = {OP}^{2} + {PQ}^{2} }} \\ \small{ \implies{ \: {(1 + PQ)}^{2} = {OP}^{2} + {PQ}^{2} }} \\ \small{ \implies{\: 1 + {PQ}^{2} + 2PQ = {OP}^{2} + {PQ}^{2} }} \\ \small{ \implies{ \: 1 + 2PQ = {7}^{2} }} \\ \small{ \implies{\: PQ = 24cm \: and \: OQ= 1 + PQ = 25cm}} \\ \small{ \implies{ \bold{So, \: sin \: Q = \frac{7}{25} \: and \: cos \: Q = \frac{24}{25}.}} }$

Answer 2

Answer:

OQ= x cm then

PQ = 1+x

by Pythagoras theorem we get

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)