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Question Number 214183 by jlewis last updated on 30/Nov/24

$$\mathrm{a})\mathrm{The}\mathrm{angle}\mathrm{of}\mathrm{elevation}\mathrm{from}\mathrm{a}\mathrm{point}\mathrm{A}\mathrm{to}\mathrm{the}$$$$\mathrm{top}\mathrm{of}\mathrm{building}5\mathrm{m}\mathrm{away}\mathrm{is}45°.\mathrm{Another}\mathrm{point}$$$$\mathrm{B}\mathrm{is}4\mathrm{m}\mathrm{from}\mathrm{A}.\mathrm{By}\mathrm{scale}\mathrm{drawing}\mathrm{determine}.$$$$\mathrm{i})\mathrm{the}\mathrm{angles}\mathrm{of}\mathrm{elevation}\mathrm{from}\mathrm{point}\mathrm{B}$$$$\mathrm{ii})\mathrm{the}\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{building}$$$$\mathrm{b})\mathrm{A}\mathrm{man}\mathrm{who}\mathrm{is}\mathrm{in}\mathrm{the}\mathrm{elevator}\mathrm{half}\mathrm{way}\mathrm{the}$$$$\mathrm{building}\mathrm{notices}\mathrm{a}\mathrm{lorry}\mathrm{approaching}\mathrm{the}$$$$\mathrm{building}\mathrm{at}\mathrm{an}\mathrm{angle}\mathrm{of}\mathrm{depression}20°.\mathrm{How}$$$$\mathrm{far}\mathrm{is}\mathrm{the}\mathrm{lorry}\mathrm{from}\mathrm{the}\mathrm{foot}\mathrm{of}\mathrm{the}\mathrm{building}?$$

Answered by MrGaster last updated on 24/Dec/24

$$\underset{\mathrm{Determine}\mathrm{height}\mathrm{h}\mathrm{from}\mathrm{point}A}{\underbrace{\tan \left(45°\right)=\frac{h}{5}}}\Rrightarrow h=5\tan \left(45°\right)=5$$$$\underset{\mathrm{Determine}\mathrm{angln}\mathrm{of}\mathrm{elevation}\mathrm{from}\mathrm{point}B}{\underbrace{\tan \left({\theta }_B\right)=\frac{h}{d_B}}}\Rrightarrow {\theta }_B=\arctan \left(\frac{5}{\sqrt{4^2+5^2}}\right)$$$$\underset{\mathrm{Distance}\mathrm{from}B\mathrm{to}\mathrm{the}\mathrm{building}}{\underbrace{d_B=\sqrt{4^2+5^2}}}\Rrightarrow d_B=\sqrt{16+25}=\sqrt{41}$$$${\theta }_B=\arctan \left(\frac{5}{\sqrt{41}}\right)$$$$\frac{h}{2}\left(\mathrm{Height}\mathrm{of}\mathrm{elevator}\right)=\frac{5}{2}$$$$\underset{\mathrm{Determine}\mathrm{distance}\mathrm{of}\mathrm{lorry}\mathrm{from}\mathrm{foot}\mathrm{of}\mathrm{building}}{\underbrace{\tan (-20°)=\frac{\frac{5}{2}}{d_L}}}\Rrightarrow d_L=\frac{\frac{5}{2}}{\tan (-{20}^{°})}$$$$d_L=\frac{\frac{5}{2}}{\tan (-20°)}$$$$d_L=\frac{5}{2\tan \left(20°\right)}$$

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