Question

You and another person are sunbathing on the beach near a lifeguard station. The other person chooses a spot that is the same distance from the shoreline but 30 feet closer to the station than you. The angles of elevation from you and the other person to the top of the lifeguard station are 26° and 51°, respectively. Estimate the height of the lifeguard station to the nearest tenth of a foot.

Answer 1

Let's label the points: A for your position, B for the other person's position, C for the shoreline point, and D for the top of the lifeguard station. We have two right triangles: ΔACD and ΔBCD.

We're given:
- Angle CAD = 26°
- Angle CBD = 51°
- Distance BC = 30 feet

Let's find the distance CD using the tangent function:

tan(51°) = height / BC
height = tan(51°) * 30

Now, let's find the distance AC using the tangent function:

tan(26°) = height / AC
height = tan(26°) * AC

Since the height of the lifeguard station is the same in both equations, we can equate them:

tan(51°) * 30 = tan(26°) * AC

Now, solve for AC:

AC = (tan(51°) * 30) / tan(26°)

Now, we can find the height of the lifeguard station using either triangle:

height = tan(26°) * AC
height ≈ 15.5 feet

So, the height of the lifeguard station is approximately 15.5 feet to the nearest tenth of a foot.