Question

A lifeguard spots a swimmer in the lake at an angle of depression of 32°

and spots a sailboat, further out, at an angle of depression of 13°. If the
lifeguard is 36.5 dm away from the swimmer and 84.7 dm away from the
sailboat, how far is the swimmer from the sailboat? Round your answer to
the nearest tenth of a dm. Include a diagram

Answer 1

To solve this problem, we can use trigonometry and set up two right triangles. Let's call the distance between the swimmer and sailboat "x". We can use the tangent function to find the distances:

tan(32°) = height of swimmer / distance to swimmer
tan(13°) = height of sailboat / distance to sailboat

Multiplying both sides of each equation by the given distances, we get:

height of swimmer = 36.5 tan(32°)
height of sailboat = 84.7 tan(13°)

We can set up a third right triangle with the height of the sailboat as one leg, the distance between the sailboat and swimmer as the other leg, and "x" as the hypotenuse. We can use the tangent function again to find "x":

tan(θ) = height of swimmer - height of sailboat / distance between swimmer and sailboat

Substituting the given values and the expressions we found for the heights, we get:

tan(θ) = (36.5 tan(32°)) - (84.7 tan(13°)) / x

Simplifying:

x tan(θ) = (36.5 tan(32°)) - (84.7 tan(13°))
x = [(36.5 tan(32°)) - (84.7 tan(13°))] / tan(θ)

Substituting the given angle of depression for the sailboat (13°), we get:

x = [(36.5 tan(32°)) - (84.7 tan(13°))] / tan(13°)
x = 196.1 dm

Therefore, the swimmer is approximately 196.1 dm away from the sailboat.

Here's a diagram to help visualize the problem:

Answer 2

The distance between the swimmer and the sailboat is approximately 22.8 dm. This is determined using trigonometry, considering the angles of depression and distances from the lifeguard to both the swimmer and the sailboat.

We want to find the distance between the swimmer and the sailboat SB.

Identify the relevant trigonometric relationships

We can use the tangent function for both angles:

`\[\tan(\angle AS) = \frac{{\text{{opposite side}}}}{{\text{{adjacent side}}}} \quad \text{(for angle \(AS\))}\]\[\tan(\angle AB) = \frac{{\text{{opposite side}}}}{{\text{{adjacent side}}}} \quad \text{(for angle \(AB\))}\]`

Write the equations

For angle AS:

`\[\tan(32^\circ) = \frac{{SB}}{{AS}}\]`

For angle \(AB\):

`\[\tan(13^\circ) = \frac{{SB}}{{AB}}\]`

Solve for SB

Let's solve these equations simultaneously:

`\[SB = \tan(32^\circ) * AS\]\[SB = \tan(13^\circ) * AB\]`

Using the given values:

`\[SB = \tan(32^\circ) * 36.5\]\[SB = \tan(13^\circ) * 84.7\]`

Calculate the numerical values

`\[SB \approx 22.83 \, \text{dm} \quad \text{(rounded to the nearest tenth)}\]`

So, the swimmer is approximately 22.83 dm away from the sailboat.