Question

Two men on either side of the cliff 80m high observes the angle of elevation of the top of the cliff to be 30° and 60° respectively. Find the distance between the two men.

Answer 1

Let the Cliff be denoted by : CF

As the Men are either side of the Cliff (CF) :

✿ Let the Man on the Left side of the Cliff be M(1)

✿ Let the Distance between M(1) and Cliff be D(1)

✿ Let the Man on the Right side of the Cliff be M(2)

✿ Let the Distance between M(2) and Cliff be D(2)

Given :

✿ Angle of Elevation of the Top of the Cliff with respect to M(1) is 30°

✿ Angle of Elevation of the Top of the Cliff with respect to M(2) is 60°

We know that :

✿ $\sf{Tan\theta = \dfrac{Opposite\;Side}{Adjacent\;Side}}$

From the Figure :

✿  $\sf{Tan60 = \dfrac{Height\;of\;the\;Cliff}{Distance\;Between\;Man(1)\;and\;Cliff}}$

$\implies \sf{\sqrt{3} = \dfrac{80}{D(1)}$

$\implies \sf{D(1) = \dfrac{80}{\sqrt{3}}$

✿  $\sf{Tan30 = \dfrac{Height\;of\;the\;Cliff}{Distance\;Between\;Man(2)\;and\;Cliff}}$

$\implies \sf{\dfrac{1}{\sqrt{3}} = \dfrac{80}{D(2)}$

$\implies \sf{D(2) = 80\sqrt{3}}$

We can notice that, The Distance between the two men is the Sum of the Distances D(1) and D(2)

$\sf{\implies Distance\;between\;the\;Two\;Men : D(1) + D(2)}$

$\sf{\implies Distance\;between\;the\;Two\;Men : \dfrac{80}{\sqrt{3}} + 80\sqrt{3}}$

$\sf{\implies Distance\;between\;the\;Two\;Men : 80(\dfrac{1}{\sqrt{3}} + \sqrt{3})}$

$\sf{\implies Distance\;between\;the\;Two\;Men : 80\times (\dfrac{4}{\sqrt{3}})}$

$\sf{\implies Distance\;between\;the\;Two\;Men : \dfrac{320}{\sqrt{3}}}$

$\sf{\implies Distance\;between\;the\;Two\;Men : 184.76\;Meters}$

Answer 2

$<b>$

Given,
Height of the cliff = 80 mts

Assume that the two men are at Points C and D and are observing the elevation angle of the cliff.

As per the given data,
$\angle{ADB} = 30^{\circ}$
$\angle{ACB} = 60^{\circ}$

AB represents the cliff Height
CB + BD represents the distance between the two men

Here, We have 2 Right angled triangles viz.,
$\triangleABD$ and $\triangleACB$

$tan = \frac{Opposite}{Adjacent}$

From $\triangle{ABD}$
$tan\: 30^{\circ} = \frac{AB}{BD}$

=> $\frac{1}{\sqrt{3}}= \frac{80}{BD}$

=> $BD = 80\sqrt{3}\: mts$ ---------[1]

From $\triangle{ABC}$
$tan\: 60^{\circ} = \frac{AB}{CB}$

=> $\sqrt{3}= \frac{80}{CB}$

=> $CB = \frac{80}{\sqrt{3}}\: mts$ ---------[2]

We have to Add [1]&[2] ,
Simplify [1] to take $80\sqrt{3}$ as a common term

=> $CB = \frac{80}{\sqrt{3}}\: ×\frac{\sqrt{3}}{\sqrt{3}}$

=> $CB = \frac{80\sqrt{3}}{3}$ ----------[3]

Add [1] & [3]
$BD + CB = \frac{80\sqrt{3}}{3}+ 80\sqrt{3}$

=>$BD + CB = \sqrt{3}(\frac{80}{3} + 80)$

=>$BD + CB = \sqrt{3}(\frac{80+240}{3})$

=> $BD + CB = \sqrt{3}(\frac{320}{3})$

=> $BD + CB = 184.75\: mts$

Therefore ,
The distance between the two men observing the elevation of cliff = 184.75 mts