Question

a hot air balloon is 300 m high. the pilot sees the starting point at an angle of depression of 25°. how far away is the balloon from the starting point along the ground? round your answer to the nearest hundredth. question 6 1 pts find the angle of elevation to a 60 - ft tower when the observer stands 80 ft from its base. round your answers to the nearest tenth of a degree. angle of elevation: degrees

Explanation:

Step1: Analyze the first problem

We know the height of the balloon (opposite side) is \(h = 300\) m and the angle of depression \(\theta=25^{\circ}\). The angle of depression is equal to the angle of elevation from the starting - point to the balloon. Let \(x\) be the distance along the ground. We use the tangent function \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\). So \(\tan(25^{\circ})=\frac{300}{x}\).

Step2: Solve for \(x\) in the first problem

$$x=\frac{300}{\tan(25^{\circ})}$$

Since \(\tan(25^{\circ})\approx0.4663\), then \(x=\frac{300}{0.4663}\approx643.36\) m.

Step3: Analyze the second problem

We have the height of the tower (opposite side) \(y = 60\) ft and the adjacent side \(z = 80\) ft. Let \(\alpha\) be the angle of elevation. Using the tangent function \(\tan\alpha=\frac{\text{opposite}}{\text{adjacent}}\), so \(\tan\alpha=\frac{60}{80}=0.75\).

Step4: Solve for \(\alpha\) in the second problem

\(\alpha=\tan^{- 1}(0.75)\)
Using a calculator, \(\alpha\approx36.9^{\circ}\)

Answer:

For the first problem: \(643.36\)
For the second problem: \(36.9\)