\#5.) SAT Question! In a right triangle, one angle measures \( x^{\circ} \), where \( \sin x^{\circ}=\frac{4}{5} \). What is \( \cos \left(90^{\circ}-x^{\circ}\right) ? \)
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In a right triangle, the sine of an angle is equal to the cosine of its complementary angle. So, if \( \sin x^{\circ} = \frac{4}{5} \), it also holds that \( \cos(90^{\circ} - x^{\circ}) = \sin x^{\circ} \). Therefore, \( \cos(90^{\circ} - x^{\circ}) = \frac{4}{5} \). Additionally, using the Pythagorean theorem, we can find that if \( \sin x^{\circ} = \frac{4}{5} \), the opposite side is 4 and the hypotenuse is 5. The adjacent side can be calculated as \( \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3 \). Thus, \( \cos x^{\circ} = \frac{3}{5} \), reaffirming \( \cos(90^{\circ} - x^{\circ}) = \frac{4}{5} \).