Question

The shadow of a tower standing on a level ground is found to be 40m longer when Sun's altitude is 30° than when it was 60°. Find the height of the tower.

Answer 1

$<b>$

Let Tower's height be 'h' mts

When Sun's Altitude = 60°
=> $\angle{ACB} =60°$
=> Length of the Shadow = BC

When Sun's Altitude = 30°
=> $\angle{ADB} =30°$
=> Length of the Shadow = DB

Given,
When Altitude of Sun changes ,
Length of the shadow DB = Length of the Shadow BC + 40

Since ,
BD = BC + CD -------[1]

We have,
CD = 40 mts

$\angle{ABD} = 90^{\circ}$

From $\triangleABC$
$tan\: 60^{\circ}=\frac{AB}{BC}$

=> $\sqrt{3}=\frac{h}{BC}$

=> $BC =\frac{h}{\sqrt{3}}\: mts$ ------[a]

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

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

=> $BD =\sqrt{3}h\: mts$ -----[b]

From[1]
$BD = BC + CD$

$\sqrt{3}h = \frac{h}{\sqrt{3}} + 40$

=> $3h = h+ 40\sqrt{3}$

=> $2h = 40\sqrt{3}$

=> $h = 20\sqrt{3}\: mts$

Therefore,
Tower's Height is $20\sqrt{3}\: mts$

Answer 2

Hey buddy here is ur solution !!

GIVEN : Tower be AB
When sun's altitude is 60°
angle ACB = 60°
Length of shadow =BC
When sun's altitude is 30°
angle ADB = 30°
Length of Shadow = DB
Shadow is 40 m when angle change from 60° to 30°
CD = 40m
angle ABC = 90°

IN TRIANGLE ABC
TAN60° = AB/CB
ROOT3 = AB/CB
CB = AB/ROOT3 ----(1)

IN TRIANGLE ABD
TAN30° = AB/DB
1/ROOT3 = AB/DB
DB = ROOT3 AB

DC + CB = ROOT3 AB
40 + CB = ROOT3 AB
CB = ROOT3 AB - 40 -----(2)

FROM (1) & (2)

AB/ROOT3 = ROOT3 AB - 40
AB = ROOT3 (ROOT3 AB) - 40 ROOT3
AB= 3AB - AB
40 ROOT3 = 3AB -AB
40 ROOT3 = 2AB
AB = 40 ROOT3 / 2
AB = 20 ROOT3

HEIGHT OF TOWER = AB = 20 ROOT3 m

Hope u like my process

(^_^)



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