Question

The shadow of a tower standing on a level ground is found to be 40m longer when the sun's altitude is 30° than when it is 60°. Find the height of the tower.
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Answer 1

$\huge\bf\red{SOLUTION:}$

AB is the tower and BC is the length of the shadow when the sun's altitude is 60°,I.e, the angle of elevation of the top of the tower from the tip of the shadow is 60° and DB is the length of the shadow, when the angle of elevation is 30°.

Now let AB be h metre and BC be x metre according to the question DB is 40m longer than BC

so,

DB=(40+x)m

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Now, we have two right Triangles ABC and ABD:

In triangle ABC,

$tan60degree = \frac{ab}{bc} \\ \sqrt{3} = \frac{h}{x} - - - - - (1)$

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In triangle ABD,

$tan30deg = \frac{ab}{bd} \\ \frac{1}{ \sqrt{3} } = \frac{h}{x + 40} - - - - - (2)$

From (1) we have,

$h = x \sqrt{3}$

Putting this value in (2),

We get,

$(x \sqrt{3} ) \sqrt{3} = x + 40 \\ i.e \: 3x = x + 40 \\ x =20$

hence, 20root3 m is the answer

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Answer 2

Given : The difference in shadow when ∅ shifted from 60° to 30° is 40 m .

To find : Height of the tower .

Solution : Let the height of the tower be h m and the length of shadow when ∅ is 60° be x m .

In triangle ABC ,

$\sf \tan 60 {}^{o} \: = \frac{perpendicular}{base} = \frac{h}{x}$

$\sf \implies\sqrt{3} = \frac{h}{x}$

$\sf \: \implies \: x \sqrt{3} = h$

Now , in triangle ABD

$\sf \: tan \: 30 {}^{o} = \frac{h}{40 + x}$

$\sf \ \implies \: \frac{1}{ \sqrt{3} } = \frac{h}{40 + x}$

Using value of h from the above equation

$\sf \: \implies \: \frac{1}{ \sqrt{3} } = \frac{x \sqrt{3} }{40 + x}$

By cross multiplication

$\sf \: \implies \: 40 + x = x \sqrt{3} \times \sqrt{3}$

$\sf \: \implies \: 40 + x = 3x$

$\implies \sf \: 40 = 3x - x$

$\implies \sf \: 40 = 2x$

$\sf \implies \: x = \frac{40}{2}$

$\implies \sf \: x = 20 \: m$

From the above results ,

$\rm \: h = x \sqrt{3}$

$\boxed{ \rm \: h \: = 20 \sqrt{3} m} \: \\ \rm \: or \\ \rm \: h \: = 20 \times 1.73 \\ \boxed{ \rm \: h = 34.6 \: m}$

Refer to the figure in attachment