Question

using the appropriate special triangle, determine $\theta$ if $0^{circ}leq\thetaleq90^{circ}$ for $cos\theta=\frac{sqrt{3}}{2}$.

a) $90^{circ}$
b) $30^{circ}$
c) $45^{circ}$
d) $60^{circ}$

Explanation:

Step1: Recall special - triangle trigonometric values

We know the \(30^{\circ}-60^{\circ}-90^{\circ}\) triangle. In a right - angled triangle with angles \(30^{\circ},60^{\circ},90^{\circ}\), if the hypotenuse \(c = 2\), the side adjacent to \(30^{\circ}\) is \(\sqrt{3}\), and the side opposite to \(30^{\circ}\) is \(1\).
The cosine formula is \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\).

Step2: Substitute into the cosine formula

Let \(\theta\) be an acute angle. For \(\cos\theta=\frac{\sqrt{3}}{2}\), using the formula \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\), when the adjacent side \(a=\sqrt{3}\) and hypotenuse \(c = 2\), from the properties of the \(30^{\circ}-60^{\circ}-90^{\circ}\) triangle, \(\theta = 30^{\circ}\)

Answer:

B. \(30^{\circ}\)