Question Number 228009 by Mingma last updated on 09/Mar/26
Answered by fantastic2 last updated on 09/Mar/26
Commented by fantastic2 last updated on 09/Mar/26
$$\angle{OAB}=\mathrm{20}^{\mathrm{0}} \Rightarrow\angle{AOB}=\mathrm{140}^{\mathrm{0}} \Rightarrow\angle{ACB}=\mathrm{70}^{\mathrm{0}} \\ $$$$\Rightarrow\angle{DBC}=\mathrm{180}^{\mathrm{0}} −\left(\mathrm{70}^{\mathrm{0}} +\mathrm{30}^{\mathrm{0}} +\mathrm{40}^{\mathrm{0}} \right)=\mathrm{40}^{\mathrm{0}} \\ $$$$\therefore\angle{BDC}=\mathrm{180}^{\mathrm{0}} −\mathrm{40}^{\mathrm{0}} −\mathrm{70}^{\mathrm{0}} =\mathrm{70}^{\mathrm{0}} \\ $$$$\therefore\angle{BOE}=\mathrm{90}^{\mathrm{0}} −\mathrm{40}^{\mathrm{0}} −\mathrm{20}^{\mathrm{0}} =\mathrm{30}^{\mathrm{0}} \\ $$$$\therefore{OB}=\mathrm{2}{BE}={BC}={BD}={r} \\ $$$$\therefore{OB}={BD} \\ $$$$\therefore\angle{ODB}=\mathrm{80}^{\mathrm{0}} \\ $$$$ \\ $$
Commented by Mingma last updated on 09/Mar/26
Fantastic
Commented by Mingma last updated on 09/Mar/26
How do you know AE is perpendicular to CB?
Commented by fantastic2 last updated on 09/Mar/26
$${AE}\:{is}\:{not}\:{perpendicular}\:{to}\:{BC} \\ $$$${OE}\bot{BC} \\ $$
Answered by mr W last updated on 09/Mar/26
Commented by mr W last updated on 09/Mar/26
$$\angle{BEC}=\mathrm{30}+\mathrm{20}+\mathrm{20}=\mathrm{70}° \\ $$$${BOD}\:{is}\:{diameter}. \\ $$$$\angle{BCE}=\angle{BDA}=\mathrm{90}−\mathrm{20}=\mathrm{70}° \\ $$$$\angle{CBE}=\mathrm{180}−\mathrm{70}−\mathrm{70}=\mathrm{40}° \\ $$$$\angle{OBC}=\mathrm{20}+\mathrm{40}=\mathrm{60}° \\ $$$${OB}={OC}\:\Rightarrow\:{OBC}\:{is}\:{equilateral} \\ $$$${OB}={CB}={EB}\:\Rightarrow\:{BOE}\:{is}\:{isosceles} \\ $$$${X}=\frac{\mathrm{180}−\mathrm{20}}{\mathrm{2}}=\mathrm{80}° \\ $$
Commented by Mingma last updated on 09/Mar/26
Excellent!