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16. You are traveling 20 mph on a completely straight, horizontal scenic trail. You look up and see the top of a mountain at an angle of [tex]\pi / 4[/tex]. Three minutes later, you look up and the top is at an angle of [tex]60^{\circ}[/tex]. How high is the top of the mountain above the scenic trail?



Answer :

Sure, let's solve this problem step-by-step.

### Step 1: Convert Time to Hours
You are traveling at a speed of 20 mph. Given that you travel for 3 minutes, we first need to convert this time into hours because the speed given is in miles per hour (mph).

[tex]\[ \text{Time in hours} = \frac{\text{Time in minutes}}{60} = \frac{3}{60} = 0.05 \text{ hours} \][/tex]

### Step 2: Calculate the Distance Traveled in 3 Minutes
Using the time converted into hours, we can now find out how far you traveled in 3 minutes.

[tex]\[ \text{Distance traveled} = \text{Speed} \times \text{Time} = 20 \text{ mph} \times 0.05 \text{ hours} = 1 \text{ mile} \][/tex]

### Step 3: Define Variables and Given Angles
Let:
- [tex]\(d\)[/tex] be the initial distance to the base of the mountain.
- [tex]\(h\)[/tex] be the height of the mountain.

You see the top of the mountain at two different angles:
- At the initial point, [tex]\(\theta_1 = \frac{\pi}{4}\)[/tex].
- After traveling 1 mile, [tex]\(\theta_2 = 60^\circ\)[/tex].

We need to convert [tex]\(60^\circ\)[/tex] to radians since the first angle is given in radians.

[tex]\[ \theta_2 = 60^\circ = \frac{\pi}{3} \text{ radians} \][/tex]

### Step 4: Set Up the Tangent Relationships
Using trigonometry, we can set up the following relationships using the tangent function because tangent of an angle in a right triangle is the opposite side (height of the mountain) divided by the adjacent side (distance to the base):

1. Initially:
[tex]\[ \tan\left(\frac{\pi}{4}\right) = \frac{h}{d} \][/tex]

2. After traveling 1 mile:
[tex]\[ \tan\left(\frac{\pi}{3}\right) = \frac{h}{d - 1} \][/tex]

### Step 5: Solve the Tangent Equations
The values for the tangents are:
[tex]\[ \tan\left(\frac{\pi}{4}\right) = 1 \][/tex]
[tex]\[ \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \][/tex]

Let's set up the equations:
1. From the initial angle:
[tex]\[ 1 = \frac{h}{d} \implies h = d \][/tex]

2. From the second angle:
[tex]\[ \sqrt{3} = \frac{h}{d - 1} \][/tex]

Since [tex]\(h = d\)[/tex], substitute [tex]\(d\)[/tex] for [tex]\(h\)[/tex] in the second equation:
[tex]\[ \sqrt{3} = \frac{d}{d - 1} \][/tex]

### Step 6: Solve for [tex]\(d\)[/tex]
Multiply both sides by [tex]\((d - 1)\)[/tex] to avoid division:
[tex]\[ \sqrt{3}(d - 1) = d \][/tex]
[tex]\[ \sqrt{3}d - \sqrt{3} = d \][/tex]

Combine like terms:
[tex]\[ \sqrt{3}d - d = \sqrt{3} \][/tex]

Factor out [tex]\(d\)[/tex] on the left:
[tex]\[ d(\sqrt{3} - 1) = \sqrt{3} \][/tex]
[tex]\[ d = \frac{\sqrt{3}}{\sqrt{3} - 1} \][/tex]

Multiply the numerator and denominator by [tex]\((\sqrt{3} + 1)\)[/tex] to rationalize the denominator:
[tex]\[ d = \frac{\sqrt{3}(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{3 + \sqrt{3}}{3 - 1} = \frac{3 + \sqrt{3}}{2} \][/tex]

Numerically this simplifies to approximately:
[tex]\[ d \approx 2.366 \][/tex]

### Step 7: Calculate the Height [tex]\(h\)[/tex]
Given that [tex]\(h = d\)[/tex]:
[tex]\[ h \approx 2.366 \text{ miles} \][/tex]

### Final Answer
The height of the mountain above the scenic trail is approximately:
[tex]\[ h \approx 2.366 \text{ miles} \][/tex]