Question

select an appropriate function to find the angle indicated (round to 10ths of a degree). 43. 44.

Explanation:

Step1: Recall trigonometric ratio

For a right - triangle, \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\), \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\)

Problem 43

Step1: Identify the sides relative to \(\theta\)

The side opposite to \(\theta\) is \(6\) m and the side adjacent to \(\theta\) is \(18\) m.
We use the tangent function: \(\tan\theta=\frac{6}{18}\)

Step2: Solve for \(\theta\)

\(\theta=\tan^{- 1}(\frac{6}{18})\)
\(\theta=\tan^{-1}(\frac{1}{3})\)
Using a calculator, \(\theta\approx18.4^{\circ}\)

Problem 44

Step1: Find the length of the other side (opposite to \(\beta\))

Using the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), where \(c = 15\) in (hypotenuse) and \(b = 14\) in (adjacent to \(\beta\)). Let the opposite side be \(a\).
\(a=\sqrt{15^{2}-14^{2}}=\sqrt{(15 + 14)(15 - 14)}=\sqrt{29}\approx5.4\) in

Step2: Use the sine function

\(\sin\beta=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{\sqrt{29}}{15}\)
\(\beta=\sin^{-1}(\frac{\sqrt{29}}{15})\)
Using a calculator, \(\beta\approx21.3^{\circ}\)

Answer:

1. \(\theta\approx18.4^{\circ}\)
2. \(\beta\approx21.3^{\circ}\)