# Trigonometry word problem

A tower $150$ meter high is situated at the top of a hill. At a point $650$ meter down the hill the angle between the surface of the hill and the line of sight to the top of the tower is $12^\circ$ $30$ minutes. Find the inclination of the hill to a horizontal plane.

How would I solve this problem?

I made a right triangle but so I know one of the angle is $90^\circ$ the other one must $77^\circ$ degrees $30$ minutes and faces $650$ meter and the other is $12^\circ$ $30$ minutes and faces $150$ meters.

But what would I do?

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For the diagram, please refer to Ross Millikan's answer. I am stealing his picture and labels. The main difference between our answers is that in the one below, we can proceed directly by calculator.

Let $\phi =\angle CDB$. By the Sine Law for $\triangle BDC$, $$\frac{\sin\phi}{650}=\frac{\sin 12.5^\circ}{150}.$$

Now we can find $\phi$ to excellent accuracy with the calculator. I assume you know how to do that (calculate $\sin\phi$, press the $\sin^{-1}$ button). To $1$ decimal place, $\phi \approx 69.7^\circ$.

It follows that $\angle ABD$ is, to $1$ decimal place, $90^\circ-69.7\circ=20.3^\circ$. To find the angle the hill makes with the horizontal, subtract $12.5^\circ$. We get $7.8^\circ$. Convert the $0.8$ part to minutes if you wish.

Alternately, let $\theta=\angle ABD$. Then $\sin\phi=\cos\theta$, so the Sine Law gives directly $\frac{\cos\theta}{650}=\frac{\sin 12.5^\circ}{150}$.

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Great explanation and answer. –  Fernando Martinez Aug 15 '12 at 15:04

Draw a picture. Point A is inside the hill, directly below the tower. $\angle CBD = 12^\circ 30',\ CD=150, \ BC=650$. So if $\theta=\angle ABC, AC=650 \sin \theta, AB= 650 \cos \theta, AD=AC+150=AB \tan (\theta + 12^\circ 30')$ Put it all together and you have an equation for $\theta$, which I think will require numeric solution.

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I like this diagram. –  Fernando Martinez Aug 15 '12 at 14:54