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?
$\angle CBD = 12^\circ 30',\ CD=150, \ BC=650$. 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.