SAT Mathematics Practice Test 3

Since the angles are acute and$\mathit{s}\mathit{i}\mathit{n}\left(\mathit{a}°\right)=\mathit{c}\mathit{o}\mathit{s}\left(\mathit{b}°\right)$,
it follows from the complementary angle property of sines and cosines that $\mathit{a}+\mathit{b}=90$.
Substituting $4\mathit{k}-22$ for a and $6\mathit{k}-13$ for b gives $\left(4\mathit{k}-22\right)+\left(6\mathit{k}-13\right)=90$,
which simplifies to $10\mathit{k}-35=90$.
Therefore, $10\mathit{k}=125$, and $\mathit{k}=12.5.$
Choice A is incorrect and may be the result of mistakenly assuming that $\mathit{a}+\mathit{b}$ and making a sign error.
Choices B and D are incorrect because they result in values for a and b such that $\mathit{s}\mathit{i}\mathit{n}\left(\mathit{a}°\right)\ne \mathit{c}\mathit{o}\mathit{s}\left(\mathit{b}°\right)$.