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**Hint:**Draw a diagram by carefully reading the question and applying the constraints given in it. Assume two variables (say ‘x’ and ‘y’) for distance of point ‘A’ and ‘B’ respectively from the base of the tower. Then use the formula $\text{tan }\!\!\theta\!\!\text{ =}\dfrac{\text{perpendicular}}{\text{base}}$ in two triangle separately to find ‘x’ and ;y; and then calculate distance between ‘A’ and ‘B’ by substituting ‘x’ from ‘y’.

**Complete step by step answer:**

According to question, we can draw the following diagram-

We have to find the distance between points ‘A’ and ‘B’. Let us assume the distance of point ‘A’ from the base of tower to be x.m and the distance of point B from the base of tower to be ‘y’m.

From the diagram, we can see that-

Distance between points ‘A’ and ‘B’ $=\left( y-x \right)m$ …………………………….. (1)

To find the distance between ‘A’ and ‘B’, we need to find ‘x’ and ‘y’.

Let us find ‘x’ first-

In the diagram, we can see that-

$\angle NMA+\angle AMO={{90}^{\circ }}$ (Complementary angles)

Given angle of depression for point A $={{60}^{\circ }}$ .

$\Rightarrow \angle NMA={{60}^{\circ }}$

So, $\begin{align}

& {{60}^{\circ }}+\angle AMO={{90}^{\circ }} \\

& \Rightarrow \angle AMO={{90}^{\circ }}-{{60}^{\circ }} \\

& \Rightarrow \angle AMO={{30}^{\circ }} \\

\end{align}$

We know that $\text{tan }\!\!\theta\!\!\text{ =}\dfrac{\text{perpendicular}}{\text{base}}$

In \[\Delta OMA\] -

For $\angle AMO$ -

Perpendicular $=OA=xm$

Base $=OM=15m$ .

So, $\tan \angle AMO=\dfrac{OA}{OM}$ .

On putting $\angle AMO={{30}^{\circ }}$ , $OA=x.m$ and $OM=15m$ , we will get-

$\tan {{30}^{\circ }}=\dfrac{x}{15}$

We know $\tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}}$ .

On putting $\tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}}$ , we will get-

$\Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{x}{15}$ .

On multiplying both sides by 15, we will get-

$\Rightarrow \dfrac{15}{\sqrt{3}}=x$ .

On multiplying both numerator and denominator by $\sqrt{3}$ in LHS, we will get

$\begin{align}

& \Rightarrow \dfrac{15\sqrt{3}}{\sqrt{3}\times \sqrt{3}}=x \\

& \Rightarrow \dfrac{15\sqrt{3}}{3}=x \\

& \Rightarrow 5\sqrt{3}=x \\

& \Rightarrow x=5\sqrt{3}m \\

\end{align}$

Now let us find ‘y’

In the diagram, we can see that

$\angle NMB+\angle BMO={{90}^{\circ }}$ (Complementary angles)

Given angle of depression for point B$={{45}^{\circ }}$

$\Rightarrow \angle NMB={{45}^{\circ }}$

So, ${{45}^{\circ }}+\angle BMO={{90}^{\circ }}$

$\begin{align}

& \angle BMO={{90}^{\circ }}-{{45}^{\circ }} \\

& \angle BMO={{45}^{\circ }} \\

\end{align}$

Again, we will apply $\text{tan }\!\!\theta\!\!\text{ =}\dfrac{\text{perpendicular}}{\text{base}}$ for $\angle BMO$ in triangle OMB.

In $\Delta OMB$ -

Fro $\angle BMO$ -

Perpendicular $=OB=y.m$

Base $=OM=xm$ .

So, $\tan \angle BMO=\dfrac{OB}{OM}$ .

On putting $\angle BMO={{45}^{\circ }}$. $OB=y.m$ and $OM=15m$ , we will get-

$\tan {{45}^{\circ }}=\dfrac{ym}{15m}$ .

We know that $\tan {{45}^{\circ }}=1$ .

On putting $\tan {{45}^{\circ }}=1$ , we will get-

$1=\dfrac{y}{15}$ .

On multiplying both sides by 15, we will get-

$\begin{align}

& 15=y \\

& \Rightarrow y=15m \\

\end{align}$

Now, we have obtained $x=5\sqrt{3}m$ and $y=15m$ , So, putting $x=5\sqrt{3}m$ and $y=15m$ , we will get-

Distance between A and B \[=15-5\left( 1.732 \right)\]

$\begin{align}

& =15-8.660 \\

& =6.34 \\

\end{align}$

**Hence, the required distance between points A and B is 6.34m.**

**Note:**Students generally make mistakes in recognizing the angle of depression.

If from point ‘O’, we are watching point ‘A’, angle ‘∅’ is the angle of depression. But students get confused and take angle ‘α’ as an angle of depression.

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