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Question-159099




Question Number 159099 by akolade last updated on 12/Nov/21
Answered by Rasheed.Sindhi last updated on 13/Nov/21
Let h_1  is the height of tower above   kim′s position and h_2  the height below  her position.  The height of tower h=h_1 +h_2        tan 42=(h_1 /(16))⇒h_1 =16.tan 42       tan 24=(h_2 /(16))⇒h_2 =16.tan 24  h=h_1 +h_2 =16(tan 42+tan 24)                 =16(1.3456)=21.53m
$${Let}\:{h}_{\mathrm{1}} \:{is}\:{the}\:{height}\:{of}\:{tower}\:{above}\: \\ $$$${kim}'{s}\:{position}\:{and}\:{h}_{\mathrm{2}} \:{the}\:{height}\:{below} \\ $$$${her}\:{position}. \\ $$$$\mathcal{T}{he}\:{height}\:{of}\:{tower}\:{h}={h}_{\mathrm{1}} +{h}_{\mathrm{2}} \\ $$$$\:\:\:\:\:\mathrm{tan}\:\mathrm{42}=\frac{{h}_{\mathrm{1}} }{\mathrm{16}}\Rightarrow{h}_{\mathrm{1}} =\mathrm{16}.\mathrm{tan}\:\mathrm{42} \\ $$$$\:\:\:\:\:\mathrm{tan}\:\mathrm{24}=\frac{{h}_{\mathrm{2}} }{\mathrm{16}}\Rightarrow{h}_{\mathrm{2}} =\mathrm{16}.\mathrm{tan}\:\mathrm{24} \\ $$$${h}={h}_{\mathrm{1}} +{h}_{\mathrm{2}} =\mathrm{16}\left(\mathrm{tan}\:\mathrm{42}+\mathrm{tan}\:\mathrm{24}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{16}\left(\mathrm{1}.\mathrm{3456}\right)=\mathrm{21}.\mathrm{53}{m} \\ $$

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