Question

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

(a) $5(\sqrt{3}+1)\mathrm{m}$
(b) $10(\sqrt{3}-1)\mathrm{m}$
(c) 9 m
(d) 13 m

Answer 1

(a) $5(\sqrt{3}+1)\mathrm{m}$
Let $AB$ be the tower and $AC\mathrm{and}AD$ be its shadows.
Thus, we have:
∠$ACB={45}^o$ and ∠$ADB={30}^o$
If $AC=x$ m, then $AD=(x+10)$ m.
Let:
$AB\hspace{0.17em}=h$ m.

In ∆$ACB$, we have:
$\frac{AC}{AB}=cot{45}^o=1$

⇒ $\frac{x}{h}=1$
⇒ $h=x$ ... (i)
Now, in ∆$ADB$, we have:
$\frac{AD}{AB}=cot{30}^o=\sqrt{3}$
⇒ $\frac{(x+10)}{h}=\sqrt{3}$
⇒ $x+10=\sqrt{3}h$ ...(ii)
On putting the value of $h$ from (i) in (ii), we get:
$h+10=\sqrt{3}h$
⇒ $(\sqrt{3}-1)h=10$
⇒ $h=\frac{10}{(\sqrt{3}-1)}$
On multiplying the numerator and denominator by $(\sqrt{3}+1)$, we get:
$h=\frac{10}{(\sqrt{3}-1)}\times \frac{(\sqrt{3}+1)}{(\sqrt{3}+1)}=\frac{10(\sqrt{3}+1)}{3-1}=\frac{10(\sqrt{3}+1)}{2}=5(\sqrt{3}+1)$ m

Hence, the height of the tower is $5(\sqrt{3}+1)$ m.