ABC-is-a-triangular-park-with-AB-AC-100-m-A-clock-tower-is-situated-at-the-midpoint-of-BC-The-angles-of-elevation-of-top-of-the-tower-at-A-and-B-are-cot-1-3-2-and-cosec-1-2-6-respectiv

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Question Number 17647 by Tinkutara last updated on 09/Jul/17

$$ABC\mathrm{is}\mathrm{a}\mathrm{triangular}\mathrm{park}\mathrm{with}AB=$$$$AC=100\mathrm{m}.\mathrm{A}\mathrm{clock}\mathrm{tower}\mathrm{is}\mathrm{situated}$$$$\mathrm{at}\mathrm{the}\mathrm{midpoint}\mathrm{of}BC.\mathrm{The}\mathrm{angles}\mathrm{of}$$$$\mathrm{elevation}\mathrm{of}\mathrm{top}\mathrm{of}\mathrm{the}\mathrm{tower}\mathrm{at}A\mathrm{and}$$$$B\mathrm{are}{\cot }^{-1}(3.2)\mathrm{and}{\mathrm{cosec}}^{-1}(2.6)$$$$\mathrm{respectively}.\mathrm{The}\mathrm{height}\mathrm{of}\mathrm{tower}\mathrm{is}$$

Commented by ajfour last updated on 09/Jul/17

Commented by ajfour last updated on 09/Jul/17

$$\mathrm{\angle }\mathrm{AFB}=90°\mathrm{because}\mathrm{AB}=\mathrm{AC}$$$$\mathrm{and}\mathrm{F}\mathrm{is}\mathrm{midpoint}\mathrm{of}\mathrm{BC}.$$$$\mathrm{let}\mathrm{HF}=\mathrm{tower}\mathrm{height}{}^{\prime }{\mathbf{h}}^{\prime }.$$$${\mathrm{BF}}^2+{\mathrm{AF}}^2={\mathrm{AB}}^2$$$${\left(\mathrm{hcot}\theta \right)}^2+{\left(\mathrm{hcot}\varphi \right)}^2={\mathrm{AB}}^2$$$${\mathrm{h}}^2=\frac{{\mathrm{AB}}^2}{\cot {}^2\theta +\cot {}^2\varphi }$$$$=\frac{100\times {100\mathrm{m}}^2}{{(2.6)}^2-1+{(3.2)}^2}$$$$=\frac{100\times 100}{16}{\mathrm{m}}^2$$$$\Rightarrow \mathrm{h}=\frac{100}{4}\mathrm{m}=25\mathrm{m}.$$

Commented by Tinkutara last updated on 09/Jul/17

$$\mathrm{Thanks}\mathrm{Sir}!$$

Commented by ajfour last updated on 09/Jul/17

$$\mathrm{you}\mathrm{are}\mathrm{prompt}.\mathrm{good}.$$

Commented by Tinkutara last updated on 09/Jul/17

$$\mathrm{What}\mathrm{Sir}?$$

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