SATMathematics Practice Test 1

By the complementary angle relationship for sine and cosine, $\mathit{s}\mathit{i}\mathit{n}\left(\mathit{x}°\right)=\mathit{c}\mathit{o}\mathit{s}\left(90°-\mathit{x}°\right)$.
Therefore, $\mathit{c}\mathit{o}\mathit{s}\left(90°-\mathit{x}°\right)=\frac{4}{5}$ .
Either the fraction $\frac{4}{5}$ or its decimal equivalent, 0.8, may be gridded as the correct answer
Alternatively, one can construct a right triangle that has an angle of measure $\mathit{x}°$ such that $\mathit{s}\mathit{i}\mathit{n}\left(\mathit{x}°\right)=\frac{4}{5}$,
as shown in the figure below, where $\mathit{s}\mathit{i}\mathit{n}\left(\mathit{x}°\right)$ is equal to the ratio of the opposite side to the hypotenuse, or $\frac{4}{5}$
Since two of the angles of the triangle are of measure x° and 90°, the third angle must have the measure $180°-90°-\mathit{x}°=90°-\mathit{x}°$.
From the figure, $\mathit{c}\mathit{o}\mathit{s}\left(90°-\mathit{x}°\right)$, which is equal to the ratio of the adjacent side to the hypotenuse, is also $\frac{4}{5}$ .