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In the figure, sin ZMQP =__

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The question is incomplete and the figure is missing. Here is the complete question with the figure attached below.

In the figure, sin ∠MQP = ______.

A. Cos N and Sin R

B. Sin R and Sin N

C. Cos N and Sin M

D. Cos R and Sin N

Answer:

D. Cos R and Sin N

Explanation:

Given:

∠MQP = 56°

sin (∠MQP) = sin (56°)

Consider the triangle NMR,

m ∠N = 56°, m ∠R = 34°

sin (∠N) = sin (56°)

So, sin (∠MQP) = sin (∠N) = sin (56°) ---------- (1)

Now, we know that,
\sin x=\cos(90-x)

Therefore, sin (∠N) = sin (56°) = cos (90°-56°) = cos (34°)

Now, from the same triangle NMR,


\cos (\angle R)=\cos (34\°)

Therefore, sin (∠N) = cos (∠R) ------------- (2)

Hence, from equations (1) and (2), we have

sin (∠MQP) = sin (∠N) = cos (∠R)

So, option D is correct.

In the figure, sin ZMQP =__-example-1
User Chad Podoski
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