Question

A surveyor wants to find the height of a hill. He determines that the angle of elevation to the top of the hill is 50°. He then walks 40
feet farther from the base from the hill and determines that the angle of elevation to the top of the hill is now 30°. Find the height of
the hill (round to the nearest foot).
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Answer 1

The answer is 45 feet

Answer 2

Answer:

The height of  the hill is 45 feet .

Step-by-step explanation:

Refer the attached figure

Let AB be the height of hill

We are given that He determines that the angle of elevation to the top of the hill is 50°

So, $\angle ACB= 50^{\circ}$

Now  He then walks 40  feet farther from the base from the hill and determines that the angle of elevation to the top of the hill is now 30°

So, CD=40 feet

BD=BC+CD=BC+40

$\ADB= 30^{\circ}$

In ΔACB

$Tan \theta = \frac{Perpendicular}{Base}\\Tan 50^{\circ} =\frac{AB}{BC}\\1.1917 BC=AB ----1$

In ΔADB

$Tan \theta = \frac{Perpendicular}{Base}\\Tan 30^{\circ} =\frac{AB}{BD}\\\frac{1}{\sqrt{3}}=\frac{AB}{BC+40}\\\frac{1}{\sqrt{3}}(BC+40)=AB----2$

So,equate 1 and 2

$1.1917 BC=\frac{1}{\sqrt{3}}(BC+40)\\BC=37.59$

Substitute the value in equation 1

1.1917 (37.59)=AB

44.796=AB

Hence the height of  the hill is 45 feet .