from her eye, which stands 1.72 meters above the ground, myesha measures the angle of elevation to the top of a prominent skyscraper to be 19 degrees. if he is standing at a horizontal distance of 337 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

To find the height of the skyscraper, we can use the tangent of the angle of elevation. The tangent of an angle in a right-angled triangle is the ratio of the opposite side to the adjacent side.

\[ \tan(\text{angle of elevation}) = \frac{\text{height of skyscraper}}{\text{eye level height}} \]

Let \( h \) be the height of the skyscraper.

\[ \tan(19^\circ) = \frac{h}{1.72 \, \text{meters}} \]

Now, solve for \( h \):

\[ h = \tan(19^\circ) \times 1.72 \]

\[ h \approx 0.3407 \times 1.72 \]

\[ h \approx 0.5859 \]

Now, we need to find the total height, which is the height of the skyscraper plus the height of Myesha's eye level.

\[ \text{Total height} = h + 1.72 \]

\[ \text{Total height} \approx 0.5859 + 1.72 \]

\[ \text{Total height} \approx 2.3059 \]

Rounding to the nearest tenth:

\[ \text{Total height} \approx 2.3 \, \text{meters} \]

So, the height of the skyscraper is approximately 2.3 meters.