Question

For what value of x is sin x = cos 19°, where 0°< x < 90°?

A.
38°

B.
71°

C.
26°

D.
19°

Answer 1

Answer:B.  71°

Sin x = cos 19° ; 0° < x < 90°

cos is the complementary of sine. This means that the value of x in sin x and the value of 19° in cos 19° adds up to 90°

x + 19° = 90°

x = 90° - 19°

x = 71°

sin 71° = cos 19°

0.9455 = 0.9455

The value of x is 71°

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Answer 2

Step-by-step explanation:

We know that

Sin(90 - x) = cosx

Given

sinx = cos19°

and

0°<x<90°

So x is in first quadrant

Since

Sin(90 - 19)°= sin71°= Cos19°

Now

Sinx = Sin71°

x = Sin^-1Sin71°

x = 71°