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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

$$hereabcdisasquare.findthevalueof<aod$$

Answered by MJS last updated on 03/May/19

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

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