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

For what value of x is sin x = cos 19º, where 0°< x < 90°-example-1

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Check the picture below, that one in the picture is just an example of the correspondence of sine and cosine in a right-triangle, in this case, we're in the 0° < x < 90°, so in the I Quadrant, and thus we have a right-triangle.

to make it short, sin(90° - 19°) = cos(19°), or that is sin(71°), thus x = 71.

For what value of x is sin x = cos 19º, where 0°< x < 90°-example-1
User Pavlo Kyrylenko
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