Question

find one solution for the equation. assume that all angles involved are acute angles. sin(3θ - 90°)= cos(2θ - 20°) θ = ° (simplify your answer.)

Explanation:

Step1: Use co - function identity

We know that $\sin A=\cos(90^{\circ}-A)$. So, $\sin(3\theta - 90^{\circ})=\cos(90^{\circ}-(3\theta - 90^{\circ}))$.
The equation $\sin(3\theta - 90^{\circ})=\cos(2\theta - 20^{\circ})$ becomes $\cos(90^{\circ}-(3\theta - 90^{\circ}))=\cos(2\theta - 20^{\circ})$.
Simplify $90^{\circ}-(3\theta - 90^{\circ})$ to get $\cos(180^{\circ}-3\theta)=\cos(2\theta - 20^{\circ})$.
Since $\cos\alpha=\cos\beta$ implies $\alpha = 2k\pi\pm\beta,k\in\mathbb{Z}$, for acute - angle solutions, we have $180^{\circ}-3\theta=2\theta - 20^{\circ}$.

Step2: Solve for $\theta$

Add $3\theta$ to both sides of the equation $180^{\circ}-3\theta=2\theta - 20^{\circ}$:
$180^{\circ}=2\theta - 20^{\circ}+3\theta$.
Combine like terms: $180^{\circ}=5\theta - 20^{\circ}$.
Add $20^{\circ}$ to both sides: $180^{\circ}+20^{\circ}=5\theta$, so $200^{\circ}=5\theta$.
Divide both sides by 5: $\theta=\frac{200^{\circ}}{5}=40^{\circ}$.

Answer:

$40$