Can you please explain Test 3 Section 4 #23?

Yep! The key to getting this one is recognizing that when they tell you that \sin \left(a^{\circ}\right)=\cos \left(b^{\circ}\right), 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 \left(a^{\circ}\right)=\cos \left(b^{\circ}\right) and \cos \left(a^{\circ}\right)=\sin \left(b^{\circ}\right).

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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