Question

If $\sec(7q+28^\circ)=\csc(30^\circ-3q)$, then find $q$.

• A. $8^\circ$
• B. $5^\circ$
• C. $6^\circ$
• D. $9^\circ$

Hint:

Turn $\sec$/$\csc$ equations into $\cos$/$\sin$ and then use $\sin\theta=\cos(90^\circ-\theta)$ so you can apply $\cos\alpha=\cos\beta \Rightarrow \alpha=\pm\beta+360^\circ k$.

Answer 1

The Correct Option is A

Step 1: Convert to cosine equality.
$\sec A=\csc B \ \Rightarrow\ \dfrac{1}{\cos A}=\dfrac{1}{\sin B}\ \Rightarrow\ \cos A=\sin B.$
Here $A=7q+28^\circ,\ B=30^\circ-3q$. Step 2: Use $\sin\theta=\cos(90^\circ-\theta)$.
$\cos(7q+28^\circ)=\sin(30^\circ-3q)=\cos\!\big(90^\circ-(30^\circ-3q)\big)=\cos(60^\circ+3q).$ Step 3: Solve $\cos \alpha=\cos \beta$.
Either $7q+28^\circ=60^\circ+3q+360^\circ k$ or $7q+28^\circ=-(60^\circ+3q)+360^\circ k$.
From the first: $4q=32^\circ+360^\circ k \Rightarrow q=8^\circ+90^\circ k.$
Taking $k=0$ gives $q=8^\circ$ (fits options). The second branch gives no listed small positive option. \[ \boxed{8^\circ} \]