jayla542
contestada

A lifeguard is sitting on a platform, looking down at a swimmer in the water. If the lifeguard's line of
sight is 8 feet above the ground and the angle of depression to the swimmer is 18°, how far away is
the swimmer from the lifeguard?

Respuesta :

micahdisu

Answer: The swimmer is 26 ft (approximately) away from the lifeguard

Step-by-step explanation: Please refer you the attached diagram.

The lifeguard sitting on the platform is labeled as point P, and the swimmer down below is at point S. The point G is the point directly beneath the lifeguard. So, we have a right angled triangle PSG. Also the angle of depression is 18 degrees, so the angle of sight between the lifeguard and the ground is 72 degrees (that is, 90 - 18 = 72).

Having derived a right angled triangle with two sides and a reference angle, we can now use the trigonometric ratio which is;

CosP = Adjacent/Hypotenuse

The adjacent is the side that lies between the reference angle and the right angle (side PG), while the hypotenuse is the side facing the right angle (side PS). Hence;

Cos72 = 8/g

By cross multiplication we now have

g = 8/Cos72

g = 8/0.3090

g = 25.8899

Approximately, g = 26 ft.

Ver imagen micahdisu
abiolataiwo2015

How far away is the swimmer from the lifeguard is 24.62 feet.

Distance

Given:

Lifeguard=8 feet

Swimmer=18°

Hence:

Let x represent the distance between the lifeguard and the swimmer

8/x=tan18°

x=8/tan18°

x=8/0.32492

x=24.62

Inconclusion how far away is the swimmer from the lifeguard is 24.62 feet.

Learn more about distance here:https://brainly.com/question/4931057

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