Question

Two men on either side of a 80m high cliff observe 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

Two men on either sides of a 80 m high cliff observe angles of elevation on the top of a cliff.

Consider a triangle ADB such that -

• AC = height of cliff = 80 m.
• ∠ABC = 30° and ∠ADC = 60°

We have to find the distance between the two men.

In ΔACB

=> tan 30° = AC/CB

=> 1/√3 = 80/CB

=> CB = 80√3 ___ (eq 1)

In ΔACD

=> tan 60° = AC/CD

=> √3 = 80/CD

=> CD = 80/√3 ___ (eq 2)

On adding (eq 1) & (eq 2) we get,

=> CB + CD = 320/√3

Now,

Let -

• CB = BC = x
• CD = DC = y

So,

=> x + y = 320/√3

Rationalize 320/√3

=> x + y = 320/√3 × √3/√3

=> x + y = 320√3/3

=> x + y = 184.75 m

•°• Distance between two men is 184.75 m.

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 :-

$\large{\sf {\frac{CD}{DB}}}$=tan60°

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

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

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 :-

$\large{\sf {\frac{CD}{AD}}=tan30°}$

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

$\large{\boxed{\sf x = 80\sqrt{3}m}}$

━━━━━━━━━━━━━━━━━━━━━━━━━

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 :-

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