Question

The length of the shadow of a tower standing on level plane is found to be 2x metre longer when the sun's altitude is 30 than when it was 45 . Prove that the height of tower is x($\sqrt{3}$ + 1) metres .

Answer 1

$\tan {45}^{\circ }=\frac{AB}{BC}\Rightarrow 1=\frac{h}{y}\Rightarrow y=h\tan {30}^{\circ }=\frac{AB}{BD}=\frac{h}{2x+y}\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{2x+h}(\because y=h)\Rightarrow 2x+h=\sqrt{3}h\Rightarrow (\sqrt{3}-1)h=2xh=\frac{2x}{\sqrt{3}-1}=\frac{2x(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}=\frac{2(\sqrt{3}+1)x}{3-1}=\frac{2(\sqrt{3}+1)x}{2}=(\sqrt{3}+1)x$