Question

Two lifeguards at the lake are stationed 28 meters apart. They both located a struggling swimmer at the same time. The first lifeguard indicated that the position of the swimmer made an angle of 50 degrees with the line between the lifeguard chairs. The second lifeguard indicated the swimmer made an angle of 56 degrees with the same line. How far is the first lifeguard from the swimmer?

*Solve Using Law of Sines*
Answer Choices Are:
A. 24.1
B. 22.3
C. 28
D. 74

Answer 1

30.33

Explanation:

To determine the distance between the first lifeguard and the swimmer, we can use trigonometry and the concept of triangulation.

Let's label the first lifeguard as A, the second lifeguard as B, and the swimmer as S. We are given that the distance between the lifeguards, AB, is 28 meters.

Now, let's draw a diagram to visualize the situation. We have a triangle with vertices A, B, and S. The line connecting A and B represents the distance between the lifeguard chairs. The angles formed between the line AB and the lines AS and BS are 50 degrees and 56 degrees, respectively.

To find the distance between the first lifeguard (A) and the swimmer (S), we can use the Law of Sines. The Law of Sines states that the ratio of the length of a side of a triangle to the sine of the opposite angle is constant.

In this case, we can set up the following equation:

sin(50 degrees) / AB = sin(56 degrees) / AS

Now, let's solve for AS, which represents the distance between the first lifeguard and the swimmer:

AS = (AB * sin(56 degrees)) / sin(50 degrees)

Substituting the given value of AB as 28 meters, we can calculate AS as follows:

AS = (28 * sin(56 degrees)) / sin(50 degrees)

Using a calculator, we can find that sin(56 degrees) is approximately 0.829 and sin(50 degrees) is approximately 0.766.

AS = (28 * 0.829) / 0.766

AS ≈ 30.33 meters

Therefore, the distance between the first lifeguard and the swimmer is approximately 30.33 meters.