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Case Study - 3: Amrita stood near the base of a lighthouse, gazing up at its towering height. She measured the angle of elevation to the top and found it to be $60^\circ$. Then, she climbed a nearby observation deck, $40$ metres higher than her original position and noticed the angle of elevation to the top of lighthouse to be $45^\circ$. (i) If $CD$ is $h$ metres, find the distance $BD$ in terms of '$h$'. (ii) Find distance $BC$ in terms of '$h$'. (iii) (a) Find the height $CE$ of the lighthouse [Use $\sqrt{3} = 1.73$]. OR (iii) (b) Find distance $AE$, if $AC = 100$ m.
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(i) $\frac{h}{BD} = \tan 45^\circ = 1 \implies BD = h$ m. (ii) $\frac{h}{BC} = \sin 45^\circ = \frac{1}{\sqrt{2}} \implies BC = \sqrt{2}h$ m. (iii)(a) $\tan 60^\circ = \frac{EC}{AE} \implies \sqrt{3} = \frac{h + 40}{h} \implies h = 54.6$ m. $CE = 94.6$ m. (iii)(b) $\cos 60^\circ = \frac{AE}{AC} \implies \frac{1}{2} = \frac{AE}{100} \implies AE = 50$ m.