Question

A electric pole 30 m high casts a shadow of 24 m. Find the height of the telephone pole which casts a shadow of 20 m at the same time.​

Answer 1

Given : -

A electric pole of 30 meters high casts a shadow of 24 meters . A telephone pole casts a shadow of 20 meters at the same time .

Required to find : -

• Height of the telephone pole ?

Trigonometery ratio used : -

Tan x = opposite side/Adjacent side

...........(or)..............

Tan x = sin x/cos x

Solution : -

Here we are given with 2 situations ;

An electric pole of 30 meters casts a shadow of 24 meters .

A telephone pole casts a shadow of 20 meters at the same time.

We were asked to find the height of the telephone pole .

So,

Using the given data let's draw the diagram representation of the above scenario.

$\setlength{\unitlength}{20} \begin{picture}(10,5) \linethickness{1} \multiput(1,1)(8,0){2}{\line(1,0){4}}\multiput(5,1)(8,0){2}{\line(0,1){3}}\multiput(0,0)(8,0){2}{\qbezier(1,1)(,)(5,4)}\multiput(0,0)(8,0){2}{\qbezier(1.7,1)(1.9,1)(1.7,1.5)}\put(5.1,2.2){ \bf Electric \: Pole }\put(1,0.5){ \bf shadow } \put(5.1,1.7){ \bf 30 \:meters }\put(2.9,0.5){ \bf 24 \: meters }\put(2,1.2){ \bf \theta }\put(13,2.5){ \bf Telephone \: pole }\put(9,0.5){ \bf Shadow \: 20 \: meters }\put(10,1.2){ \bf \theta }\end{picture}$

So,

By the diagrams we can conclude that ;

The angle of elevation is similar in both situations .

Because, the angle of elevation is always similar at every place when the time is same .

By using AA Congruency rule we can find out the height of the telephone pole .

But, let's give more preference to the trigonometry .

This implies ;

The above 2 diagram can be drawn as ;

$\:\setlength{\unitlength}{20} \begin{picture}(5,5) \linethickness{1} \put(1,1){\line(1,0){8}}\put(7,1){\line(0,1){3.8}} \put(9,1){\line(0,1){5}} \qbezier(1,1)(,)(9,6)\qbezier(2,1)(2.2,1)(2,1.6)\put(4.2,0.1){\vector( - 1,0){3.2}}\put(5.5,0.1){\vector(1,0){3.5}}\put(0.85,0.5){ \bf A }\put(6.85,0.5){ \bf B }\put(8.85,0.5){ \bf C }\put(6.85,4.9){ \bf D }\put(8.85,6.2){ \bf E }\put(3,0.5){ \bf 20 \; m }\put(4.2,0){ \bf 24 \; m }\put(7.5,0.5){ \bf 4 \; m }\put(9.2,3){ \bf 30 \; m }\put(7.2,3){ \bf x \; m }\put(2.2,1.2){ \large{\theta} }\end{picture}$

Here,

AB = length of the shadow casted by telephone pole = 20 meters

AC = length of the shadow casted by electric pole = 24 meters

BD = Height of the telephone pole = x meters

CE = Height of the electric pole = 30 meters

According to problem ;

Consider ∆ ABD

➜ Tan x = opposite side/Adjacent side ( x = theta )

• Opposite side = BD

• Adjacent side = AB

This implies ;

➜ Tan x = BD/AB

Since, AB = 20 meters

➜ Tan x = x/20 ➜ Equation - 1

Consider this as equation - 1

Similarly,

Consider ∆ ACE

In ∆ ACE ,

➜ Tan x = opposite side/Adjacent side

• Opposite side = EC

• Adjacent side = AC

➜ Tan x = EC/AC

Since, AC = 24 meters & EC = 30 meters

➜ Tan x = 30/24 ➜ Equation - 2

Consider this as equation - 2

Since, the LHS part of both equations are equal let's equate the RHS past

➜ x/20 = 30/24

➜ 24x = 30 x 20

➜ 24x = 600

➜ x = 600/24

➜ x = 300/12

➜ x = 25 meters

Therefore,

Height of the telephone pole = x = 25 meters

Answer 2

Given ,

A electric pole of 30 m high casts a shadow of 24 m and also telephone pole casts a shadow of 20 m at the same time

It implies that ,

The angle of elevation in both cases is same

Let , the height of telephone pole be " x "

In Δ ABC ,

$\tt \mapsto \tan( \theta) = \frac{30}{24}$

Now , in Δ DEF

$\tt \mapsto \tan( \theta) = \frac{x}{20}$

Since , angle of elevation is same

Thus ,

$\tt \implies \frac{30}{24} = \frac{x}{20}$

$\tt \implies \frac{5}{4} = \frac{x}{20}$

$\tt \implies x = \frac{5 \times 20}{4}$

$\tt \implies x = 5 \times 5$

$\tt \implies x = 25 \: \: m$

Hence ,

• The hight of telephone pole is 25 m