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Question-162496




Question Number 162496 by cortano last updated on 29/Dec/21
Answered by mr W last updated on 30/Dec/21
Commented by mr W last updated on 30/Dec/21
∠DCB=90−15=75°=α  α=2β  ∠FDE=β=((75°)/2)  ∠AEG=∠FDE=((75°)/2)  ∠GAE=30°+15°=45°  x=∠AEG+∠GAE=((75°)/2)+45°=82.5°
$$\angle{DCB}=\mathrm{90}−\mathrm{15}=\mathrm{75}°=\alpha \\ $$$$\alpha=\mathrm{2}\beta \\ $$$$\angle{FDE}=\beta=\frac{\mathrm{75}°}{\mathrm{2}} \\ $$$$\angle{AEG}=\angle{FDE}=\frac{\mathrm{75}°}{\mathrm{2}} \\ $$$$\angle{GAE}=\mathrm{30}°+\mathrm{15}°=\mathrm{45}° \\ $$$${x}=\angle{AEG}+\angle{GAE}=\frac{\mathrm{75}°}{\mathrm{2}}+\mathrm{45}°=\mathrm{82}.\mathrm{5}° \\ $$
Commented by Tawa11 last updated on 30/Dec/21
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$

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