Question

question 30 (1 point) using the appropriate special triangle, determine θ if 0° ≤ θ ≤ 90° for 2√2 sinθ = 2. a) 60° b) 45° c) 90° d) 30°

Explanation:

Step1: Solve for \(\sin\theta\)

Given \(2\sqrt{2}\sin\theta = 2\), divide both sides by \(2\sqrt{2}\). So, \(\sin\theta=\frac{2}{2\sqrt{2}}=\frac{1}{\sqrt{2}}\).

Step2: Recall special - triangle values

We know that for a \(45^{\circ}-45^{\circ}-90^{\circ}\) special triangle, \(\sin45^{\circ}=\frac{1}{\sqrt{2}}\).

Answer:

B. \(45^{\circ}\)