From her eye, which stands 1.69 meters above the ground, Penelope measures the angle of elevation to the top of a prominent skyscraper to be 16∘. If she is standing at a horizontal distance of 200 meters from the base of the skyscraper, what is the height of the skyscraper? Round your answer to the nearest hundredth of a meter if necessary.

You look straight parallel to ground. But when you have to watch something high, then you take your sight up by moving your head up. The angle from horizontal to the point where you stopped your head is called angle of elevation. See the figure where angle DCE is the angle when Penelope saw the skyscraper. It is measured from the horizontal line DC.

How to use right triangles to find the height of the specified sky scrapper?

Remember that we assume that building is vertical to the ground. This means, there is 90° formation.

Since the distance from the base of the building is given and the angle of elevation is given, we have enough data to construct a specific right triangle where the vertical line opposite to the angle of elevation is of the measure of the height of the building.

And then we can use tangent ratio which relates perpendicular and base to the considered angle.

Remember that trigonometric ratios are defined for right triangles only(those triangles who have one of the angle as of 90 degrees).

See the diagram attached below, where the length AE is the building's height. AD is same as CB which is height of Penelope(a person who is watching the top of that given skyscraper).

For the right angle triangle EDC, and for the angle considered ∠ECD, we have the perpendicular as side opposite to the angle ∠ECD which is ED and the base is DC. Thus, the tangent ratio of ∠ECD is given by

[tex]tan(\angle ECD) = \dfrac{Perpendicular \: to \: \angle ECD}{Base \: to \: \angle ECD } = \dfrac{ED}{DC} = \dfrac{h - 1.69}{200}\\tan(16^\circ) = \dfrac{h - 1.69}{200}\\[/tex]

DC = AB and DA = CB because of ABCD being a rectangle because of building and person both standing vertical to the ground and the line DC being parallel to the ground, thus all angles being right angle(of 90 degrees).

From the calculator, the value of tan(16°) is approx 0.2867

Thus, putting this value in the obtained equation, we get:

[tex]tan(16^\circ) = \dfrac{h - 1.69}{200}\\0.2867 = \dfrac{h-1.69}{200}\\\\0.2867 \times 200 = h - 1.69 \\57.35 + 1.69 = h\\h = 59.04 \: \rm meters[/tex]

Thus, the height of the specified skyscraper is approx 59.04 meters.

Learn more about trigonometric ratios here:

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