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Question Number 214183 by jlewis last updated on 30/Nov/24
a) The angle of elevation from a point A to the top of building 5m away is 45°. Another point B is 4m from A. By scale drawing determine. i) the angles of elevation from point B ii)the height of the building b) A man who is in the elevator half way the building notices a lorry approaching the building at an angle of depression 20°. How far is the lorry from the foot of the building?
$$\left.\mathrm{a}\right)\:\mathrm{The}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{elevation}\:\mathrm{from}\:\mathrm{a}\:\mathrm{point}\:\mathrm{A}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{top}\:\mathrm{of}\:\mathrm{building}\:\mathrm{5m}\:\mathrm{away}\:\mathrm{is}\:\mathrm{45}°.\:\mathrm{Another}\:\mathrm{point} \\ $$$$\mathrm{B}\:\mathrm{is}\:\mathrm{4m}\:\mathrm{from}\:\mathrm{A}.\:\mathrm{By}\:\mathrm{scale}\:\mathrm{drawing}\:\mathrm{determine}. \\ $$$$\left.\mathrm{i}\right)\:\mathrm{the}\:\mathrm{angles}\:\mathrm{of}\:\mathrm{elevation}\:\mathrm{from}\:\mathrm{point}\:\mathrm{B} \\ $$$$\left.\mathrm{ii}\right)\mathrm{the}\:\mathrm{height}\:\mathrm{of}\:\mathrm{the}\:\mathrm{building} \\ $$$$\left.\mathrm{b}\right)\:\mathrm{A}\:\mathrm{man}\:\mathrm{who}\:\mathrm{is}\:\mathrm{in}\:\mathrm{the}\:\mathrm{elevator}\:\mathrm{half}\:\mathrm{way}\:\mathrm{the} \\ $$$$\mathrm{building}\:\mathrm{notices}\:\mathrm{a}\:\mathrm{lorry}\:\mathrm{approaching}\:\mathrm{the} \\ $$$$\mathrm{building}\:\mathrm{at}\:\mathrm{an}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{depression}\:\mathrm{20}°.\:\mathrm{How} \\ $$$$\mathrm{far}\:\mathrm{is}\:\mathrm{the}\:\mathrm{lorry}\:\mathrm{from}\:\mathrm{the}\:\mathrm{foot}\:\mathrm{of}\:\mathrm{the}\:\mathrm{building}? \\ $$
Answered by MrGaster last updated on 24/Dec/24
tan(45°)=(h/5)_(Determine height h from point A) ⇛h=5 tan(45°)=5 tan(θ_B )=(h/d_B )_(Determine angln of elevation from point B) ⇛θ_B =arctan((5/( (√(4^2 +5^2 ))))) d_B =(√(4^2 +5^2 ))_(Distance from B to the building) ⇛d_B =(√(16+25))=(√(41)) θ_B =arctan((5/( (√(41))))) (h/2)(Height of elevator)=(5/2) tan(−20°)=((5/2)/d_L )_(Determine distance of lorry from foot of building) ⇛d_L =((5/2)/(tan(−20^° ))) d_L =((5/2)/(tan(−20°))) d_L =(5/(2 tan(20°)))
$$\underset{\mathrm{Determine}\:\mathrm{height}\:\mathrm{h}\:\mathrm{from}\:\mathrm{point}\:{A}} {\underbrace{\mathrm{tan}\left(\mathrm{45}°\right)=\frac{{h}}{\mathrm{5}}}}\:\Rrightarrow{h}=\mathrm{5}\:\mathrm{tan}\left(\mathrm{45}°\right)=\mathrm{5} \\ $$$$\underset{\mathrm{Determine}\:\mathrm{angln}\:\mathrm{of}\:\mathrm{elevation}\:\mathrm{from}\:\mathrm{point}\:{B}} {\underbrace{\mathrm{tan}\left(\theta_{{B}} \right)=\frac{{h}}{{d}_{{B}} }}}\:\Rrightarrow\theta_{{B}} =\mathrm{arctan}\left(\frac{\mathrm{5}}{\:\sqrt{\mathrm{4}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} }}\right) \\ $$$$\underset{\mathrm{Distance}\:\mathrm{from}\:{B}\:\mathrm{to}\:\mathrm{the}\:\mathrm{building}} {\underbrace{{d}_{{B}} =\sqrt{\mathrm{4}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} }}}\Rrightarrow{d}_{{B}} =\sqrt{\mathrm{16}+\mathrm{25}}=\sqrt{\mathrm{41}} \\ $$$$\theta_{{B}} =\mathrm{arctan}\left(\frac{\mathrm{5}}{\:\sqrt{\mathrm{41}}}\right) \\ $$$$\frac{{h}}{\mathrm{2}}\left(\mathrm{Height}\:\mathrm{of}\:\mathrm{elevator}\right)=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\underset{\mathrm{Determine}\:\mathrm{distance}\:\mathrm{of}\:\mathrm{lorry}\:\mathrm{from}\:\mathrm{foot}\:\mathrm{of}\:\mathrm{building}} {\underbrace{\mathrm{tan}\left(−\mathrm{20}°\right)=\frac{\frac{\mathrm{5}}{\mathrm{2}}}{{d}_{{L}} }}}\:\Rrightarrow{d}_{{L}} =\frac{\frac{\mathrm{5}}{\mathrm{2}}}{\mathrm{tan}\left(−\mathrm{20}^{°} \right)} \\ $$$${d}_{{L}} =\frac{\frac{\mathrm{5}}{\mathrm{2}}}{\mathrm{tan}\left(−\mathrm{20}°\right)} \\ $$$${d}_{{L}} =\frac{\mathrm{5}}{\mathrm{2}\:\mathrm{tan}\left(\mathrm{20}°\right)} \\ $$

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