Could you repost answer explanation to CollegeBoard Test 3, Math 4.23? Specifically, is there a way to solve this if you don’t know the little trig ID about complementary angles?

Sure. I wish this never got deleted in the first place, but I was able to find this on archive.org so below is this paragraph is the original post. As for whether this question is possible without knowing that trig fact, I think your best bet if you don’t know the fact in the moment is to recognize that there needs to be some relationship between the two angles, and although they’re not drawn to scale, I think it’s still safe to assume they’re 1) not the same angle, and 2) acute, as shown.

Yep! The key to getting this one is recognizing that when they tell you that \sin (a^\circ)=\cos (b^\circ), they’re telling you that a+b=90. That’s just a little trig fact you should keep in mind when you’re taking the SAT. It can be derived pretty easily—just picture any right triangle with acute angles measuring a^\circ and b^\circ:

Of course, since it’s a right triangle, a+b=90. And if you do your SOH-CAH-TOA, you’ll see that \sin (a^\circ )=\cos (b^\circ ) and \cos (a^\circ )=\sin (b^\circ ).

Anyway, back to the question. Once we recognize that a+b=90, the rest is just substitution.

    \begin{align*}a+b&=90\\(4k-22)+(6k-13)&=90\\10k-35&=90\\10k&=125\\k&=12.5\end{align*}

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