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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