Question

From her eye, which stands 1.73 meters above the ground, London measures the angle of elevation to the top of a prominent skyscraper to be 30^{\circ} ∘ . If she is standing at a horizontal distance of 389 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.

Answer 1

The height of the skyscraper is 226.32 meters (rounded to the nearest hundredth of a meter)

An illustrative diagram is shown in the attachment below:

In the diagram, /TB/ is the height of the skyscraper

/BL/ is the horizontal distance from the base of the skyscraper to where London is standing

Point E is her eye level

From the diagram, we observe that the height of skyscraper

/TB/ = /TM/ + /MB/ = x + 1.73m

Now, to determine the height of the skyscraper, we will first determine the value of x

Consider ΔTME

∠TEM = 30°

/TM/ is the opposite and

/ME/ is the adjacent

From the formula

$tan \theta =\frac{opposite}{adjacent}$

/TM/ = x

/ME/ 389 m

Then, we can write that

$tan30^{o} = \frac{x}{389}$

∴ x = 389×tan30°

x = 224.58925 m

Recall that the height of the skyscraper /TB/ =  x + 1.73m

∴ /TB/ =  224.58925m + 1.73m

/TB/ = 226.31925m

/TB/ ≅ 226.32m

Hence, the height of the skyscraper is 226.32 meters (rounded to the nearest hundredth of a meter)

Learn more here: brainly.com/question/22062913