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The shadow of a vertical tower on level ground increases by $10$ m when the altitude of the sun changes from the angle of elevation $45^{0}$ to $30^{0}$ . Find the height of the tower correct to one place of decimal. (take $\sqrt{3} = 1.732$ )

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13.67 m

15 m

18.67 m

20 m

answer is A.

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Detailed Solution

It is given that the shadow of a vertical tower on level ground increases by

10

m when the altitude of the sun changes from the angle of elevation

45^{0}

30^{0} .

We have to find the height of the tower .
We use

tanθ = \frac{perpendicular}{base}

.
Let us assume that the height of tower

CD

h

Now in right-angled

ΔBCD

\frac{h}{x} = \tan 45^{o}

\Rightarrow \frac{h}{x} = 1

[since

\tan 45^{o}

=1]

\Rightarrow h = x

......(i)
Again, in right-angled

ΔACD

\frac{h}{x + 10} = \tan 30^{o}

\Rightarrow \frac{h}{x + 10} = \frac{1}{\sqrt{3}}

[since

\tan 30^{o} = \frac{1}{\sqrt{3}}

]

\Rightarrow \sqrt{3} h = x + 10

\Rightarrow \sqrt{3} h = h + 10

......using (i)

\Rightarrow h (\sqrt{3} - 1) = 10

\Rightarrow h = \frac{10}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1}

\Rightarrow h = \frac{10 (\sqrt{3} + 1)}{{(\sqrt{3})}^{2} - {(1)}^{2}}

[since

(a + b) (a - b) = a^{2} - b^{2}

]

\Rightarrow h = \frac{10 (1.732 + 1)}{3 - 1}

\Rightarrow h = 5 \times 2.732

\Rightarrow h = 13.67

m
Hence, the height of the tower is

13.67

m.
Therefore, the correct option is 1.

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The shadow of a vertical tower on level ground increases by 10 m when the altitude of the sun changes from the angle of elevation 450to 300. Find the height of the tower correct to one place of decimal. (take 3=1.732)

Detailed Solution

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The shadow of a vertical tower on level ground increases by $10$ m when the altitude of the sun changes from the angle of elevation $45^{0}$ to $30^{0}$ . Find the height of the tower correct to one place of decimal. (take $\sqrt{3} = 1.732$ )