Question

question
from his eye, which stands 1.5 meters above the ground, moossa measures the angle of elevation to the top of a skyscraper to be 49°. if he is standing at a horizontal distance of 179 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.

Explanation:

Step1: Define the vertical height segment

Let $h$ be the vertical distance from Moussa's eye to the top of the skyscraper. We use the tangent function for the right triangle:
$\tan(49^\circ) = \frac{h}{179}$

Step2: Solve for $h$

Rearrange to isolate $h$:
$h = 179 \times \tan(49^\circ)$
Calculate $\tan(49^\circ) \approx 1.1504$, so:
$h \approx 179 \times 1.1504 = 205.9216$

Step3: Find total skyscraper height

Add the height of Moussa's eye above ground:
$\text{Total Height} = h + 1.5$
$\text{Total Height} \approx 205.9216 + 1.5 = 207.4216$

Step4: Round to nearest hundredth

Round the result to two decimal places:
$\text{Total Height} \approx 207.42$

Answer:

207.42 meters