Question

find an angle θ that makes the statement true. cot(3θ - 12°)=tan(θ + 14°) choose the correct answer below. a. 12° b. -78° c. 24° d. 78° e. 68° f. 90° g. 66° h. 22°

Explanation:

Step1: Use co - function identity

We know that \(\cot\alpha=\tan(90^{\circ}-\alpha)\). So, \(\cot(3\theta - 12^{\circ})=\tan(90^{\circ}-(3\theta - 12^{\circ}))\).
The given equation is \(\cot(3\theta - 12^{\circ})=\tan(\theta + 14^{\circ})\). Then \(\tan(90^{\circ}-(3\theta - 12^{\circ}))=\tan(\theta + 14^{\circ})\).
Since \(\tan A=\tan B\) implies \(A = B+180^{\circ}n\), \(n\in\mathbb{Z}\). For the principal value (assuming \(n = 0\)), we have \(90^{\circ}-(3\theta - 12^{\circ})=\theta + 14^{\circ}\).

Step2: Simplify the equation

Expand the left - hand side: \(90^{\circ}-3\theta+12^{\circ}=\theta + 14^{\circ}\).
Combine like terms: \(102^{\circ}-3\theta=\theta + 14^{\circ}\).
Add \(3\theta\) to both sides: \(102^{\circ}=4\theta + 14^{\circ}\).
Subtract \(14^{\circ}\) from both sides: \(4\theta=102^{\circ}-14^{\circ}=88^{\circ}\).
Divide both sides by \(4\): \(\theta = 22^{\circ}\).

Answer:

H. \(22^{\circ}\)