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: Use the co-function rule \( \sec\theta=\csc(90^\circ-\theta) \), so \( \sec(7q+28^\circ)=\csc\big(90^\circ-(7q+28^\circ)\big) \). Setting this equal to \( \csc(30^\circ-3q) \) gives \( 90^\circ-(7q+28^\circ)=30^\circ-3q \).

Step 2: Simplify: \( 62^\circ-7q=30^\circ-3q \Rightarrow 62^\circ-30^\circ=7q-3q \Rightarrow 32^\circ=4q \).

Step 3: Solve for \(q\): \[ q=8^\circ. \] \[ \boxed{q=8^\circ} \]