Trigonometric problem: Elevation angle

Question

The elevation of the top of a tower $KT$ from a point $A$ is $27^\circ$. At another point $B$, $50$ meters nearer to the foot of the tower where $ABK$ is a straight line, the angle of elevation is $40^\circ$. Find the height of the tower $KT$.

Answer 1

Consider the following diagram:

Looking at the outer (right-angled) triangle ($TAK$), and using trigonometry, we have: $$(1) \tan(27)=\frac{h}{50+x}.$$

Looking at the inner (right-angled) triangle ($TBK$), and using trigonometry, we have: $$(2) \tan(40)=\frac{h}{x}.$$

Now we've got a pair of simultaneous equations, $(1)$ and $(2)$. Solve them! Also bear in mind that $\tan(27)$ and $\tan(40)$ are just numbers which you can find on a calculator.

Answer 2

Draw a right angled triangle $\;\Delta AKT\;,\;\;\angle K=90^\circ\;,\;\;\angle A=27^\circ$

We also have the right angled triangle $\;\Delta BKT\;\;\;\angle KBT=40^\circ\;,\;AB=50$

If we put $\;AK=x\;$, then $\;BJ=x-50\;$ , so by trigonometry:

$$\begin{align*}\text{On}\;\;\Delta AKT:&\;\;\tan 27^\circ=\frac{KT}x\\{}\\\text{On}\;\;\Delta BKT:&\;\;\tan40^\circ=\frac{KT}{x-50}\end{align*}$$

use now both those non-linear equations in $\;x\,,\;KT\;$ to find $\;KT\;$ (and you can find $\;x\;$ as well if you want to)

Answer 3

Let $AK=d$, $BK=d-50$, and $KT=h$. Then $$ \tan 27^\circ=\frac{h}{d}\qquad\rightarrow\qquad d=\frac{h}{\tan 27^\circ}\tag1 $$ and $$ \tan 40^\circ=\frac{h}{d-50}\qquad\rightarrow\qquad d-50=\frac{h}{\tan 40^\circ}\tag2 $$ Substitute $(1)$ to $(2)$, yield \begin{align} \frac{h}{\tan 27^\circ}-50&=\frac{h}{\tan 40^\circ}\\ \frac{h}{\tan 27^\circ}-\frac{h}{\tan 40^\circ}&=50\\ h\left(\frac{1}{\tan 27^\circ}-\frac{1}{\tan 40^\circ}\right)&=50\\ h\left(\frac{\tan 40^\circ-\tan 27^\circ}{\tan 27^\circ\tan 40^\circ}\right)&=50\\ h&=50\left(\frac{\tan 27^\circ\tan 40^\circ}{\tan 40^\circ-\tan 27^\circ}\right) \end{align}