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Question Number 58979 by jahanara@gmail.com last updated on 02/May/19

Commented by jahanara@gmail.com last updated on 02/May/19

here abcd is a square.find the value of <aod

$${here}\:{abcd}\:{is}\:{a}\:{square}.{find}\:{the}\:{value}\:{of}\:<{aod} \\ $$

Answered by MJS last updated on 03/May/19

∠ADE=α the angles of the green triangle are ∠E_(green) =(α/2); ∠A=90°−((3α)/2) and 90°+α the angles of the triangle AEB are ∠A=90°−((3α)/2); ∠B=90°+(α/2); ∠E=α putting the side of the square =1 we get using (a/(sin α))=(b/(sin β))=(c/(sin γ)) in the green triangle: AE=2cos (α/2) in the triangle AEB: AE=((cos (α/2))/(sin α)) ⇒ 2cos (α/2) =((cos (α/2))/(sin α)) ⇒ sin α =(1/2) ⇒ α=30° ⇒ ∠AOD=120°

$$\angle{ADE}=\alpha \\ $$$$\mathrm{the}\:\mathrm{angles}\:\mathrm{of}\:\mathrm{the}\:\mathrm{green}\:\mathrm{triangle}\:\mathrm{are} \\ $$$$\angle{E}_{{green}} =\frac{\alpha}{\mathrm{2}};\:\angle{A}=\mathrm{90}°−\frac{\mathrm{3}\alpha}{\mathrm{2}}\:\mathrm{and}\:\mathrm{90}°+\alpha \\ $$$$\mathrm{the}\:\mathrm{angles}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}\:{AEB}\:\mathrm{are} \\ $$$$\angle{A}=\mathrm{90}°−\frac{\mathrm{3}\alpha}{\mathrm{2}};\:\angle{B}=\mathrm{90}°+\frac{\alpha}{\mathrm{2}};\:\angle{E}=\alpha \\ $$$$\mathrm{putting}\:\mathrm{the}\:\mathrm{side}\:\mathrm{of}\:\mathrm{the}\:\mathrm{square}\:=\mathrm{1} \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{using}\:\frac{{a}}{\mathrm{sin}\:\alpha}=\frac{{b}}{\mathrm{sin}\:\beta}=\frac{{c}}{\mathrm{sin}\:\gamma} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{green}\:\mathrm{triangle}:\:{AE}=\mathrm{2cos}\:\frac{\alpha}{\mathrm{2}} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{triangle}\:{AEB}:\:{AE}=\frac{\mathrm{cos}\:\frac{\alpha}{\mathrm{2}}}{\mathrm{sin}\:\alpha} \\ $$$$\Rightarrow \\ $$$$\mathrm{2cos}\:\frac{\alpha}{\mathrm{2}}\:=\frac{\mathrm{cos}\:\frac{\alpha}{\mathrm{2}}}{\mathrm{sin}\:\alpha} \\ $$$$\Rightarrow\:\mathrm{sin}\:\alpha\:=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:\alpha=\mathrm{30}° \\ $$$$\Rightarrow\:\angle{AOD}=\mathrm{120}° \\ $$

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