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Please answer q 16. I reached the cot ? tan point but can't solve it further. Please explain in brackets side by side.

**16.** The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complimentary. Prove that the height of the tower is 6 m.

Please answer q 16. I reached the cot ? tan point but cant solve it further. Please explain in brackets side by side.

$Letheightofthetowerbehm.\phantom{\rule{0ex}{0ex}}Given,theanglesofelevationofthetopoftowerfromthetwopointsarecomplementary.\phantom{\rule{0ex}{0ex}}\therefore \angle ACB=\theta and\angle ADB=90-\theta \phantom{\rule{0ex}{0ex}}In\u2206ABC,$

$In\u2206ABD,$

$\Rightarrow {\mathrm{tan}}^{2}\theta =\frac{9}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\theta =\sqrt{\frac{9}{4}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\theta =\frac{3}{2}--(neglectingthe-vevalue)$

$\therefore Heightofthetower=h=4\mathrm{tan}\theta =4\times \frac{3}{2}=2\times 3=6m(U\mathrm{sin}g(1\left)\right)\phantom{\rule{0ex}{0ex}}Thus,theheightofthetoweris6m.$

Regards

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