(iii) (a) Find the height CE of the lighthouse. (Use $\sqrt{3} = 1.73$)

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Step 1: $	an 60° = \frac{CE}{AE} \Rightarrow \sqrt{3} = \frac{h + 40}{h}$ (since $AE = BD = h$ and $CE = CD + DE = h + 40$)
Step 2: $\sqrt{3}\,h = h + 40 \Rightarrow h(\sqrt{3} - 1) = 40 \Rightarrow h = \frac{40}{\sqrt{3} - 1} = 20(\sqrt{3} + 1) = 20 	imes 2.73 = 54.6$ m. $	herefore CE = 54.6 + 40 = 94.6$ m.

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