Question

5.2 part 2
score: 9/10 answered: 9/10
question 10
the angle of elevation to the top of a building is found to be 15° from the ground at a distance of 1
mile from the base of the building. find the height of the building to the nearest foot.
feet
enter your answer as an integer or decimal number. examples: 3, -4, 5.5172
enter dne for does not exist, oo for infinity

Explanation:

Step1: Convert miles to feet

We know that \(1\) mile \( = 5280\) feet. Let the height of the building be \(h\). We have a right - triangle where the adjacent side \(x = 5280\) feet and the angle of elevation \(\theta=15^{\circ}\).

Step2: Use the tangent function

The tangent of an angle in a right - triangle is given by \(\tan\theta=\frac{opposite}{adjacent}\). Here, \(\tan\theta=\tan(15^{\circ})\) and the opposite side is the height \(h\) of the building, and the adjacent side \(x = 5280\) feet.

We know that \(\tan(15^{\circ})=\tan(45^{\circ}- 30^{\circ})=\frac{\tan45^{\circ}-\tan30^{\circ}}{1 + \tan45^{\circ}\tan30^{\circ}}=\frac{1-\frac{\sqrt{3}}{3}}{1 + 1\times\frac{\sqrt{3}}{3}}=\frac{3-\sqrt{3}}{3 + \sqrt{3}}=\frac{(3 - \sqrt{3})(3-\sqrt{3})}{(3 + \sqrt{3})(3-\sqrt{3})}=\frac{9-6\sqrt{3}+3}{9 - 3}=\frac{12-6\sqrt{3}}{6}=2-\sqrt{3}\approx0.2679\)

Since \(\tan\theta=\frac{h}{x}\), then \(h=x\tan\theta\). Substituting \(x = 5280\) and \(\tan\theta=\tan(15^{\circ})\approx0.2679\)

\(h = 5280\times0.2679\)

\(h=5280\times(0.2 + 0.06+0.0079)=5280\times0.2+5280\times0.06 + 5280\times0.0079\)

\(h = 1056+316.8+41.712\)

\(h=1414.512\approx1415\)

Answer:

\(1415\)