Question

From the foot of a hill the angle of elevation of the top of a tower is found to be $45°$. After walking $2$ km upwards along the slope of the hill which is inclined at $30°$, the same is found to be $60°$. Find the height of the tower.

Answer 1

Step 1. Draw the diagram that represents the situation of the problem

Let $BQ$ represent the height of the tower.

$\angle ACB=45°$ is the angle of elevation from the foot of the hill.

$\angle RCP=30°$ and $\angle BPQ=60°$

$PC=2km$,

Let $BQ=h,AQ=h\text{'},AR=b_{2}andCR=b_1$.

Step 2. Find the relation between line $AB$ and line $AC$.

The sum of angles of any triangle is $180°$.

An isosceles triangle has two equal sides and the angles opposite to these sides are equal.

In triangle $ACB$,

$$\begin{gathered} TanC=\frac{sideoppositetoC}{sideadjacenttoC} \\ \Rightarrow \tan \left(45\right)=\frac{AB}{BC} \\ \Rightarrow 1=\frac{AB}{BC} \\ \Rightarrow AB=BC \end{gathered}$$

Step 3. Find the value of line $PR$.

Sine function is the ratio of opposite side to hypotenuse.

In triangle $PCR$

$$\begin{gathered} \sin \theta =\frac{opposite}{hypotenuse} \\ \Rightarrow \sin 30°=\frac{PR}{PC} \\ \Rightarrow \sin 30°=\frac{h\text{'}}{2} \\ \Rightarrow \frac{1}{2}=\frac{h\text{'}}{2}[as\sin 30°=\frac{1}{2}] \end{gathered}$$

$\Rightarrow h\text{'}=1km$

Step 4. Find the value of line $RC$

In triangle $PCR$

$$\begin{gathered} \cos \theta =\frac{adjacent}{hypotenuse} \\ \Rightarrow \cos 30°=\frac{b_1}{2} \\ \Rightarrow \frac{\sqrt{3}}{2}=\frac{b_1}{2}[as\cos 30°=\frac{\sqrt{3}}{2}] \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{2\sqrt{3}}{2}=b_1 \\ \Rightarrow b_1=\sqrt{3}km \end{gathered}$$

Step 5. Find the relation between $h$ and $b_2$

In triangle $BPQ$

$$\begin{gathered} \Rightarrow \tan 60°=\frac{h}{b_2} \\ \Rightarrow \sqrt{3}=\frac{h}{b_2}[as\tan 60°=\sqrt{3}] \end{gathered}$$

$\Rightarrow h=b_2\sqrt{3}$

Step 6. Find the height of the tower

Using the above calculations, we get

$h=(AC-b_1)\sqrt{3}$ [as $b_1+b_2=AC$]

$$\begin{gathered} h=(h\text{'}+h-\sqrt{3})\sqrt{3}[asAC=AB,AB=h\text{'}+hand{b}_1=\sqrt{3}km] \\ h=(1+h-\sqrt{3})\sqrt{3}[ash\text{'}=1km] \\ h=\sqrt{3}+\sqrt{3}h-3 \\ h-\sqrt{3}h=\sqrt{3}-3 \end{gathered}$$

$$\begin{gathered} -h(\sqrt{3}-1)=-\sqrt{3}(\sqrt{3}-1) \\ h=\sqrt{3}=1.732km \end{gathered}$$

Hence, the height of the tower is $1.732km$