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A vertical tower standing on a levelled field is mounted with a vertical flag staff of length $3 \mathrm{~m}$. From a point on the field. the angles of elevation of the bottom and tip of the flag staff are $30^{\circ}$ and $45^{\circ}$ respectively. Which one of the following gives the best approximation to the height of the tower?
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The correct answer is:
$4.10 \mathrm{~m}$
as $\angle \mathrm{CPA}=45^{\circ}$
so $\mathrm{AC}=\mathrm{AP}=\mathrm{x}+3$
$\tan 30^{\circ}=\frac{\mathrm{AB}}{\mathrm{AP}}=\frac{\mathrm{x}}{\mathrm{x}+3}$
$\frac{1}{\sqrt{3}}=\frac{\mathrm{x}}{\mathrm{x}+3}$
$x+3=\sqrt{3} x$
$\mathrm{x}=\frac{3}{\sqrt{3}-1} \times \frac{(\sqrt{3}+1)}{(\sqrt{3}+1)}$
$\mathrm{x}=\frac{3 \times 2.73}{2}=\frac{8.19}{2}=4.095 \mathrm{~m} \approx 4 . \mathrm{lm}$
so $\mathrm{AC}=\mathrm{AP}=\mathrm{x}+3$
$\tan 30^{\circ}=\frac{\mathrm{AB}}{\mathrm{AP}}=\frac{\mathrm{x}}{\mathrm{x}+3}$
$\frac{1}{\sqrt{3}}=\frac{\mathrm{x}}{\mathrm{x}+3}$
$x+3=\sqrt{3} x$
$\mathrm{x}=\frac{3}{\sqrt{3}-1} \times \frac{(\sqrt{3}+1)}{(\sqrt{3}+1)}$
$\mathrm{x}=\frac{3 \times 2.73}{2}=\frac{8.19}{2}=4.095 \mathrm{~m} \approx 4 . \mathrm{lm}$
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