Question

question 32 (1 point)
using the appropriate special triangle, determine θ if 0° ≤ θ ≤ 90° for 2√2 sinθ = 2.

a) 90°
b) 45°
c) 60°
d) 30°

Explanation:

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

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

Step2: Recall special - angle sine values

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

Answer:

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