For what value of x is sin x = cos 19°, where 0°< x < 90°?
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The value of x if sin x = cos 19°, where must be 0°< x < 90° is x = 71°.
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Answer: The required value of x is 71°.
Step-by-step explanation: We are given to find he value of x for which
\(\sin x=\cos 19^\circ,\) 0°< x < 90°.
We know that
the sine of any acute angle is equal to the cosine of its complement.
So, from the given equation, we can say that
the acute angles with measure x and 19° are complement of each other.
Therefore, we get
\(x+19^\circ=90^\circ\\\\\Rightarrow x=90^\circ-19^\circ\\\\\Rightarrow x=71^\circ.\)
Thus, the required value of x is 71°.