a lifeguard sits on the lifeguard stand that is about eight feet high.he suddenly notices that a swimmer is struggling in the water.the angle between the base of the lifeguard stand and the swimmer is about 102 degree,and the straight line distance between the base of the stand and the swimmer is about 85ft. how far is the lifeguard from the swimmer?

I can't figure out the 102° angle. Describe the scene, labeling each point of interest and relevant distances and angles.

an angle between two points makes no sense.

To find the distance between the lifeguard and the swimmer, we can use trigonometry. In this case, we'll use the tangent function.

First, let's label the relevant points.

The base of the lifeguard stand is point A.
The lifeguard is point B.
The swimmer is point C.

We are given that angle BAC (the angle between the base of the lifeguard stand and the swimmer) is approximately 102 degrees, and the straight line distance AB (distance between the base of the stand and the swimmer) is about 85 ft.

Now, we can use the tangent function to find the distance BC (the distance between the lifeguard and the swimmer).

Tan(BAC) = BC / AB

First, we need to convert the angle from degrees to radians since trigonometric functions typically work with radians.

Angle BAC in radians = 102 degrees * (π / 180)

Now we can plug in the values:

Tan(102 * (π / 180)) = BC / 85

To find BC, we'll rearrange the equation:

BC = 85 * tan(102 * (π / 180))

Calculating this, we find that BC is approximately 534.25 ft.

Therefore, the lifeguard is approximately 534.25 ft away from the swimmer.