Question

Two pillars of equal heights stand on either side of a road which is 100 m wide. At a point on the road between the pillars, the angles of elevation of the tops of the pillars are ${60}^{\circ }$ and ${30}^{\circ }$. Find the height of each pillar and position of the point on the road. [Take $\sqrt{3}$=1.732] [3 MARKS]

Answer 1

Let AB and CD be two pillars, each of height h metres and let AC be the road such that AC = 100 m.

Let O be the point of observation on AC.

Let OA = x metres and OC = (100 - x) m

Also, $\angle AOB={60}^{\circ }$ and $\angle COD={30}^{\circ }$

$AB\perp AC$ and $CD\perp AC$

From right $\Delta OAB$, we have [$\frac{1}{2}$ MARK]



$\frac{AB}{OA}=tan{60}^{\circ }=\sqrt{3}$

$\Rightarrow \frac{h}{x}=\sqrt{3}\Rightarrow h=\sqrt{3}x$

In $\Delta$ ADE [$\frac{1}{2}$ MARK]

$tan60=\frac{h}{x}$

$x\sqrt{3}=h$

$x=\frac{h}{\sqrt{3}}\dots \dots \left(1\right)$ [1 MARK]

From right $\Delta OCD$, we have

$\frac{CD}{OC}=tan{30}^{\circ }=\frac{1}{\sqrt{3}}$

$\Rightarrow \frac{h}{(100-x)}=\frac{1}{\sqrt{3}}\Rightarrow h=\frac{(100-x)}{\sqrt{3}}$ .....(ii)

$\sqrt{3}x=\frac{(100-x)}{\sqrt{3}}\Rightarrow 3x=(100-x)$

$\Rightarrow 4x=100\Rightarrow x=25$

Putting x = 25 m in (i), we get

$h=(25\times \sqrt{3})=(25\times 1.732)=43.3$

In $\Delta$ BCE

$tan30=\frac{h}{100-x}$

$\frac{1}{\sqrt{3}}=\frac{h}{100-\frac{h}{\sqrt{3}}}$ from (1) [1 MARK]

$\frac{1}{\sqrt{3}}=\frac{\sqrt{3}h}{100\sqrt{3}-h}$

$100\sqrt{3}-h=3h$

$100\sqrt{3}=4h$

$h=25\left(1.732\right)$

$=43.3m$ [$\frac{1}{2}$ MARK]

Hence, the height of each pillar is 43.3 m and the point of observation is 25 m away from the first pillar.