The line sketch of the movement and positioning of Joe and the hill are shown below.
Brief description of the sketch made.
From the sketch, point A is where Joe started from heading towards the hill.
The hill is represented by BC and the height is h
25 degrees was the initial angle of elevation.
After moving 350 ft closer to the hill, the angle of elevation increased by 14 degrees to 25 + 14 = 39 degrees.
From triangle BCD,
Using the trigonometric function of tan,
[tex]\begin{gathered} \tan =\frac{opposite}{\text{adjacent}} \\ \tan 39=\frac{h}{x} \\ 0.8098=\frac{h}{x} \\ \text{Making h subject of the formula,} \\ h=0.8098x \end{gathered}[/tex]From triangle ABC,
Using the trigonometric function of tan,
[tex]\begin{gathered} \tan =\frac{opposite}{\text{adjacent}} \\ \tan 25=\frac{h}{350+x} \\ \text{Substituting the value of h gotten from the previous triangle,} \\ \tan 25=\frac{0.8098x}{350+x} \\ 0.4663=\frac{0.8098x}{350+x} \\ C\text{ ross multiplying,} \\ 0.4663(350+x)=0.8098x \\ 163.205+0.4663x=0.8098x \\ C\text{ ollecting the like terms,} \\ 163.205=0.8098x-0.4663x \\ 163.205=0.3435x \\ \text{Dividing both sides by 0.3435 to get x,} \\ x=\frac{163.205}{0.3435} \\ x=475.124ft \\ \\ \text{The height, h of the hill is;} \\ h=0.8098x \\ h=0.8098\times475.124 \\ h=384.755ft \\ h\approx385ft \end{gathered}[/tex]Therefore, the height of the hill is 385 feet
The correct answer is option B.
