Question

From his eye, which stands 1.64 meters above the ground, Keshawn measures the angle of elevation to the top of a prominent skyscraper to be 33 degrees. If he is standing at a horizontal distance of 385 meters from the base of the skyscraper, what is the height of the skyscraper? Round your answer to the nearest tenth of a meter if necessary.

Answer 1

The height of the scyscraper will be equal to H=251.66 meters

What is trigonometry?

The branch of mathematics set up a relatioship between the angle and the sides of the right angle triangle is called as trigonometry.

Here we have the following data:-

The height of the observer's eye=1.63 meters

The angle which eye is making withe the building=33

The horizontal distance = 385 meters

So the height of the building is calculated as:-

$Tan\theta=\dfrac{P}{B}\\\\\\tan\theta=\dfrac{H-1.64}{385}\\\\ \\\\tan33=\dfrac{H-1.64}{385}$

H=251.66 meters

hence height of the scyscraper will be equal to $H=251.66\ meters$

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