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. \[ \boxed{8^\circ} \]

Answer 2

This equation can be solved quickly using the co-function relationship between secant and cosecant, without first converting everything to cosine form.

Setting up the identity. Since \( \csc(90^\circ - \theta) = \dfrac{1}{\sin(90^\circ-\theta)} = \dfrac{1}{\cos\theta} = \sec\theta \), the equation \( \sec A = \csc B \) is equivalent to \( A + B = 90^\circ \) (taking the principal solution). Here \( A = 7q+28^\circ \) and \( B = 30^\circ - 3q \), so \[ (7q+28^\circ) + (30^\circ-3q) = 90^\circ. \] This simplifies to \( 4q + 58^\circ = 90^\circ \), giving \( 4q = 32^\circ \) and \( q = 8^\circ \).

Checking each option directly against \( A+B=90^\circ \) confirms this without ambiguity:

1. Option \(8^\circ\): \(A = 7(8)+28 = 84^\circ\), \(B = 30-3(8) = 6^\circ\), and \(A+B = 90^\circ\) exactly — this value satisfies the equation.
2. Option \(5^\circ\): \(A = 35+28=63^\circ\), \(B=30-15=15^\circ\), \(A+B=78^\circ \neq 90^\circ\) — does not satisfy the equation.
3. Option \(6^\circ\): \(A=42+28=70^\circ\), \(B=30-18=12^\circ\), \(A+B=82^\circ \neq 90^\circ\) — does not satisfy the equation.
4. Option \(9^\circ\): \(A=63+28=91^\circ\), \(B=30-27=3^\circ\), \(A+B=94^\circ \neq 90^\circ\) — does not satisfy the equation.

Only \(q=8^\circ\) makes the two angles complementary, which is exactly the condition needed for \(\sec A=\csc B\).

So the correct answer is \(8^\circ\).