Question

If the elevation of the sun changes form 30° to 60°, then the difference between the lengths of shadows of a pole 15 m high, is

(a) 7.5 m
(b) 15 m
(c) $10\sqrt{3}\mathrm{m}$
(d) $5\sqrt{3}\mathrm{m}$

Answer 1

(c) $10\sqrt{3}\mathrm{m}$
Let $AB$ be the pole and $AC$ and $AD$ be its shadows.
We have:
∠$ACB={30}^o$, ∠$ADB={60}^o$ and $AB=15$m

In ∆$ACB$, we have:
$\frac{AC}{AB}=cot{30}^o=\sqrt{3}$
⇒ $\frac{AC}{15}=\sqrt{3}$ ⇒ $AC=15\sqrt{3}$ m

Now, in ∆$ADB$, we have:
$\frac{AD}{AB}=cot{60}^o=\frac{1}{\sqrt{3}}$
⇒ $\frac{AD}{15}=\frac{1}{\sqrt{3}}$ ⇒ $AD=\frac{15}{\sqrt{3}}$ $=\frac{15\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}=\frac{15\sqrt{3}}{3}=5\sqrt{3}$ m

∴ Difference between the lengths of the shadows = $AC-AD=15\sqrt{3}-5\sqrt{3}=10\sqrt{3}$ m