Question

find one solution for the equation. assume that all angles involved are acute angles. sec (3β + 42°) = csc (β − 8°) β = □° (simplify your answer.)

Explanation:

Step1: Recall reciprocal and co-function identities

We know that \(\sec\theta=\csc(90^{\circ}-\theta)\) and also if \(\sec A = \csc B\), then \(A + B=90^{\circ}\) (since \(\sec A=\csc(90 - A)\) and if \(\csc(90 - A)=\csc B\), for acute angles \(90 - A=B\) or \(A + B = 90^{\circ}\)).

Given \(\sec(3\beta+ 42^{\circ})=\csc(\beta - 8^{\circ})\), using the co - function identity \(\sec x=\csc(90^{\circ}-x)\), we can rewrite the equation as:

\(\csc(90^{\circ}-(3\beta + 42^{\circ}))=\csc(\beta-8^{\circ})\)

Since the cosecant function is one - to - one for acute angles (the range of acute angles is \(0^{\circ}<\theta<90^{\circ}\)), we can set the arguments equal:

\(90^{\circ}-(3\beta + 42^{\circ})=\beta-8^{\circ}\)

Step2: Simplify the left - hand side of the equation

Simplify \(90^{\circ}-(3\beta + 42^{\circ})\):

\(90^{\circ}-3\beta-42^{\circ}=48^{\circ}-3\beta\)

So our equation becomes \(48^{\circ}-3\beta=\beta - 8^{\circ}\)

Step3: Solve for \(\beta\)

Add \(3\beta\) to both sides of the equation:

\(48^{\circ}=\beta-8^{\circ}+3\beta\)

\(48^{\circ}=4\beta-8^{\circ}\)

Add \(8^{\circ}\) to both sides:

\(48^{\circ}+8^{\circ}=4\beta\)

\(56^{\circ}=4\beta\)

Divide both sides by 4:

\(\beta=\frac{56^{\circ}}{4} = 14^{\circ}\)

We need to check if the angles are acute.

For \(3\beta + 42^{\circ}\), substitute \(\beta = 14^{\circ}\): \(3\times14^{\circ}+42^{\circ}=42^{\circ}+42^{\circ}=84^{\circ}\) (acute, since \(84^{\circ}<90^{\circ}\))

For \(\beta-8^{\circ}\), substitute \(\beta = 14^{\circ}\): \(14^{\circ}-8^{\circ}=6^{\circ}\) (acute, since \(6^{\circ}<90^{\circ}\))

Answer:

\(14\)