Question

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

Answer 1

Answer:

The distance between men at point A and men at point B is $\dfrac{320}{\sqrt{3} }$ meters  

Step-by-step explanation:

Given as :

Two men on either side of cliff observes the angles of elevation of the top of the cliff to be 30° and 60° respectively .

let The two men are at point C and B

The distance of point C from tower base = AC = x meters

The distance of point B from tower base = AB = y meters

The height of cliff = h = 80 meters

Let The distance between men at point A and men at point B = (x + y) meters

The height of cliff = OA = h = 80 meters

According to question

From figure

In Δ OAC

Tan angle = $\dfrac{perpendicular}{base}$

Or, Tan 30° = $\dfrac{OA}{AC}$

Or, $\dfrac{1}{\sqrt{3} }$ = $\dfrac{h}{x}$

∴   x = √3 h  

i.e  x = 80√3    

So, The distance of point C from tower base = AC = x =  80√3 meters    

Again

In Δ OAB

Tan angle = $\dfrac{perpendicular}{base}$

Or, Tan 60° = $\dfrac{OA}{AB}$

Or, √3 = $\dfrac{h}{y}$

∴   y =  $\dfrac{h}{\sqrt{3} }$            

i.e  y = $\dfrac{80}{\sqrt{3} }$

So, The distance of point B from tower base = AB = y = $\dfrac{80}{\sqrt{3} }$ meters  

∴, The distance between men at point A and men at point B = (x+ y) meters

i.e x + y = (80√3 + $\dfrac{80}{\sqrt{3} }$ ) meters

Or,  x + y = 80 ( $\dfrac{3+1}{\sqrt{3} }$ )

∴  x + y = $\dfrac{320}{\sqrt{3} }$ meters

So, The distance between men at point A and men at point B =  x + y = $\dfrac{320}{\sqrt{3} }$ meters

Hence, The distance between men at point A and men at point B is $\dfrac{320}{\sqrt{3} }$ meters   . Answer

Answer 2

$\huge\underline\purple{\sf Answer:-}$

$\large{\boxed{\sf s = 184.75m}}$

$\huge\underline\purple{\sf Solution:-}$

Given :-

Hight of cliff :- 80 m

Angle of elevation :- 30° and 60°

To find

Distance between two men = ?

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Let,

A and B be the position between 2 men .

Distance between AD = x

And Distance between DB = y

In ∆ BCD :-

$\large{\sf {\frac{CD}{DB} = tan \: 60°}}$

$\large\implies{\sf {\frac{80}{y}}=\sqrt3}$

$\large\implies{\sf y ={\frac{80}{\sqrt{3}m}}}$

On rationalisation :-

$\large\implies{\sf {\frac{80}{\sqrt{3}}}×{\frac{\sqrt{3}}{\sqrt{3}}}}$

$\large{\boxed{\sf y={\frac{80✓3}{3}}m}}$

In ∆ ACD :-

$\large{\sf {\frac{CD}{AD}}=tan30°}$

$\large\implies{\sf {\frac{80}{x}}={\frac{1}{\sqrt{3}}}}$

$\large{\boxed{\sf x = 80\sqrt{3}m}}</p><p>$

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Distance Between two men :-

Distance Between Two men(s) = x + y

$\large\implies{\sf {\frac{80\sqrt{3}}{3}}+80\sqrt{3}}$

On solving this we get :-

$\huge\red{\boxed{\sf s = 184.75m}}$