From his eye, which stands 1.63 meters above the ground, Isaac measures the angle of elevation to the top of a prominent skyscraper to be $17^{\circ}$ . If he is standing at a horizontal distance of 294 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

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Answer

$h \approx 91.46$

Steps

Step 1 ： $t a n (17^{\circ}) = \frac{h - 1.63}{294}$

Step 2 ： $h = 1.63 + 294 \cdot t a n (17^{\circ})$

Step 3 ： $h \approx 91.46$

Question

From his eye, which stands 1.63 meters above the ground, Isaac measures the angle of elevation to the top of a prominent skyscraper to be $17^{\circ}$ . If he is standing at a horizontal distance of 294 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.

Expert–verified

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Answer

$h \approx 91.46$

Steps

Step 1 ： $t a n (17^{\circ}) = \frac{h - 1.63}{294}$

Step 2 ： $h = 1.63 + 294 \cdot t a n (17^{\circ})$

Step 3 ： $h \approx 91.46$

Accepted Answer

Step:1 ：(tan(17^{circ}) = frac{h - 1.63}{294})Step:2 ：(h = 1.63 + 294 cdot tan(17^{circ}))Step:3 ：(h approx 91.46)

Answer

Step: 1 ：(tan(17^{circ}) = frac{h - 1.63}{294})Step: 2 ：(h = 1.63 + 294 cdot tan(17^{circ}))Step: 3 ：(h approx 91.46)