Question : The length of the shadow of a vertical tower on level ground increases by 10 m when the altitude of the sun changes from 45° to 30°. The height of the tower is:
Option 1: $10 \sqrt{3}$ m
Option 2: $5 \sqrt{3}$ m
Option 3: $5(\sqrt{3}+1)$ m
Option 4: $10(\sqrt{3}+1)$ m
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Correct Answer: $5(\sqrt{3}+1)$ m
Solution :
Let the height of the tower be $h$ meters.
And $p$ is the shadow of the tower
In $\triangle ABC$,
⇒ $\tan45^\circ = \frac{h}{p}$
⇒ $p = h$....................................(equation i)
In $\triangle ABD$
⇒ $\tan30^\circ = \frac{h}{10+p}$
Putting the value $p$ from equation (i), we get:
⇒ $\frac{1}{\sqrt3} = \frac{h}{h+10}$
⇒ $10 + h = \sqrt3h$
⇒ $h(\sqrt3 - 1) = 10$
⇒ $h = \frac{10}{(\sqrt3 - 1)}$
On rationalisation
⇒ $h = \frac{10(\sqrt3 + 1)}{(\sqrt3 - 1)(\sqrt3 + 1)}$
⇒ $h = \frac{10(\sqrt3 + 1)}{2}$
⇒ h= $5(\sqrt{3}+1)$ m
Hence, the correct answer is $5(\sqrt{3}+1)$ m.
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