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From-the-top-of-a-tower-80m-high-a-student-observe-that-the-angle-of-depression-of-the-top-of-a-vertical-pole-A-is-45-and-the-angle-of-depression-of-point-B-on-the-pole-is-60-if-O-is-the-foot-of

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Question Number 9813 by tawakalitu last updated on 05/Jan/17
From the top of a tower 80m high, a student observe that the angle of depression of the top of a vertical pole A is 45° and the angle of depression of point B on the pole is 60°. if O is the foot of the pole and OB = 20m. Find the distance between A and B. leaving your answer in surd form.
$$\mathrm{From}\:\mathrm{the}\:\mathrm{top}\:\mathrm{of}\:\mathrm{a}\:\mathrm{tower}\:\mathrm{80m}\:\mathrm{high},\:\mathrm{a}\:\mathrm{student} \\ $$$$\mathrm{observe}\:\mathrm{that}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{depression}\:\mathrm{of}\:\mathrm{the}\:\mathrm{top}\: \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{vertical}\:\mathrm{pole}\:\mathrm{A}\:\mathrm{is}\:\mathrm{45}°\:\mathrm{and}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{of}\: \\ $$$$\mathrm{depression}\:\mathrm{of}\:\mathrm{point}\:\mathrm{B}\:\mathrm{on}\:\mathrm{the}\:\mathrm{pole}\:\mathrm{is}\:\mathrm{60}°. \\ $$$$\mathrm{if}\:\mathrm{O}\:\mathrm{is}\:\mathrm{the}\:\mathrm{foot}\:\mathrm{of}\:\mathrm{the}\:\mathrm{pole}\:\mathrm{and}\:\mathrm{OB}\:=\:\mathrm{20m}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}. \\ $$$$\mathrm{leaving}\:\mathrm{your}\:\mathrm{answer}\:\mathrm{in}\:\mathrm{surd}\:\mathrm{form}. \\ $$
Answered by sandy_suhendra last updated on 05/Jan/17
Commented by sandy_suhendra last updated on 05/Jan/17
∠BPQ=60° tan 60° = ((BQ)/(PQ)) (√3) = ((60)/(PQ)) ⇒ PQ = ((60)/( (√3))) = 20(√3) m ∠APQ=45° tan 45° = ((AQ)/(PQ)) 1 = ((AQ)/(20(√3))) ⇒AQ = 20(√3) m AB=BQ−AQ=(60−20(√3)) m
$$\angle\mathrm{BPQ}=\mathrm{60}° \\ $$$$\mathrm{tan}\:\mathrm{60}°\:=\:\frac{\mathrm{BQ}}{\mathrm{PQ}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\sqrt{\mathrm{3}}\:=\:\frac{\mathrm{60}}{\mathrm{PQ}}\:\Rightarrow\:\mathrm{PQ}\:=\:\frac{\mathrm{60}}{\:\sqrt{\mathrm{3}}}\:=\:\mathrm{20}\sqrt{\mathrm{3}}\:\mathrm{m} \\ $$$$ \\ $$$$\angle\mathrm{APQ}=\mathrm{45}° \\ $$$$\mathrm{tan}\:\mathrm{45}°\:=\:\frac{\mathrm{AQ}}{\mathrm{PQ}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:=\:\frac{\mathrm{AQ}}{\mathrm{20}\sqrt{\mathrm{3}}}\:\Rightarrow\mathrm{AQ}\:=\:\mathrm{20}\sqrt{\mathrm{3}}\:\mathrm{m} \\ $$$$ \\ $$$$\mathrm{AB}=\mathrm{BQ}−\mathrm{AQ}=\left(\mathrm{60}−\mathrm{20}\sqrt{\mathrm{3}}\right)\:\mathrm{m} \\ $$
Commented by tawakalitu last updated on 05/Jan/17
Wow. God bless you sir.
$$\mathrm{Wow}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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