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

Solution:

Given, sin x = cos 19°, where 0° < x < 90°

We have to find the value of x.

We know, 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°

So, x + 19° = 90°

x = 90° - 19°

x = 71°

Verification:

Sin 71° = cos 19°

LHS = sin 71° = 0.9455

RHS = cos 19° = 0.9455

LHS = RHS

Therefore, the value of x is 71°.

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

Summary:

The value of x in sin x = cos 19°, where 0° < x < 90°is 71°.