# Trigonometry - heights and distances

The angle of elevation of a cliff from a fixed point A is $\theta$. After going up a distance of k metres towards the top of the cliff at an angle of $\phi$ it is found that the angle of elevation is $\alpha$. Show that the height of the cliff in metres is $k(\cos\phi -\sin\phi\cot\alpha)/(\cot \theta- \cot \alpha)$.

Let $h$ be the height from $A$, and let $d$ be the horizontal distance from $A$ to the base of the cliff. Then $$\frac{h}{d}=\tan\theta.$$ When we walked a total distance $k$ at an angle $\phi$, we gained $k\sin\phi$ in altitude, so the height from where we are is now $h-k\sin\phi$. Our horizontal distance from the base decreased by $k\cos\phi$, so is now $d-k\cos\phi$. It follows that $$\frac{h-k\sin\phi}{d-k\cos \phi}=\tan\alpha.$$ We want to eliminate $d$. The first equation can be written as $d=h\cot\theta$, and the second as $d=(h-k\sin\phi)\cot\alpha +k\cos\phi$. Thus $$h\cot\theta=h\cot \alpha-k\sin\phi\cot\alpha +k\cos\phi.$$ This is a linear equation in $h$. Solve. We get the desired expression.