Question

Choose the correct answer of the following question:

The length of the shadow of a tower standing on level ground is found to be 2x metres longer when the sun's elevation is 30° than when it was
45°. The height of the tower is

(a) $\left(2\sqrt{3}x\right)$ m (b) $\left(3\sqrt{2}x\right)$ m (c) $\left(\sqrt{3}-1\right)x$ m (d) $\left(\sqrt{3}+1\right)x$ m

Answer 1

Let CD = h be the height of the tower.

We have,

$$\begin{gathered} \mathrm{AB}=2x,\angle \mathrm{DAC}=30°\mathrm{and}\angle \mathrm{DBC}=45° \\ \mathrm{In}∆\mathrm{BCD}, \\ \tan 45°=\frac{\mathrm{CD}}{\mathrm{BC}} \\ \Rightarrow 1=\frac{h}{\mathrm{BC}} \\ \Rightarrow \mathrm{BC}=h \\ \mathrm{Now},\mathrm{in}∆\mathrm{ACD}, \\ \tan 30°=\frac{\mathrm{CD}}{\mathrm{AC}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{\mathrm{AB}+\mathrm{BC}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{2x+h} \\ \Rightarrow 2x+h=h\sqrt{3} \\ \Rightarrow h\sqrt{3}-h=2x \\ \Rightarrow h\left(\sqrt{3}-1\right)=2x \\ \Rightarrow h=\frac{2x}{\left(\sqrt{3}-1\right)}\times \frac{\left(\sqrt{3}+1\right)}{\left(\sqrt{3}+1\right)} \\ \Rightarrow h=\frac{2x\left(\sqrt{3}+1\right)}{\left(3-1\right)} \\ \Rightarrow h=\frac{2x\left(\sqrt{3}+1\right)}{2} \\ \therefore h=x\left(\sqrt{3}+1\right)\mathrm{m} \end{gathered}$$

Hence, the correct answer is option (d).