etch and Solve each problem given below. Round your answers to two decimal places. 1.) The angle of elevation from point $$ A $$ to the top of a hill is $$ 49^{\circ} $$. If point $$ A $$ is 400 feet from the base of the hill, how high is the hill? 2.) Find the angle of elevation of the sun when a 12.5 -meter-tall telephone pole casts a 18 -meter- long shadow.

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1.) We are given the angle of elevation $$ \theta = 49^{\circ} $$ and the horizontal distance $$ d = 400 $$ feet. For a right triangle, the tangent of the angle is the ratio of the opposite side (hill height $$ h $$) to the adjacent side (distance $$ d $$). Thus, we have

$$
\tan(49^{\circ}) = \frac{h}{400}.
$$

Solving for $$ h $$:

$$
h = 400 \cdot \tan(49^{\circ}).
$$

Using a calculator for $$ \tan(49^{\circ}) $$:

$$
\tan(49^{\circ}) \approx 1.15.
$$

Thus,

$$
h \approx 400 \cdot 1.15 \approx 460.16 \text{ feet}.
$$

2.) We are given the telephone pole height $$ H = 12.5 $$ meters and the length of the shadow $$ s = 18 $$ meters. The angle of elevation $$ \theta $$ is given by the relationship:

$$
\tan(\theta) = \frac{H}{s} = \frac{12.5}{18}.
$$

Calculating the quotient:

$$
\frac{12.5}{18} \approx 0.69.
$$

To find the angle $$ \theta $$, take the inverse tangent:

$$
\theta = \arctan(0.69).
$$

Using a calculator:

$$
\theta \approx 34.99^{\circ}.
$$

Thus, the angle of elevation of the sun is approximately $$ 34.99^{\circ} $$.
1.) The hill is approximately 460.16 feet high. 2.) The angle of elevation of the sun is approximately 34.99 degrees.

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