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Question Number 117481 by I want to learn more last updated on 12/Oct/20

Answered by mr W last updated on 12/Oct/20

$$letAC=1$$$$\frac{OC}{\sin 20}=\frac{AC}{\sin (10+20)}$$$$\Rightarrow OC=\frac{\sin 20}{\sin 30}=2\sin 20$$$$\frac{BC}{\sin 40}=\frac{AC}{\sin (30+10+20+20)}$$$$\Rightarrow BC=\frac{1}{2\cos 40}$$$$\frac{BC}{\sin (x+30)}=\frac{OC}{\sin x}$$$$\frac{\sin (x+30)}{\sin x}=\frac{BC}{OC}=\frac{1}{4\cos 40\sin 20}$$$$\cos 30+\frac{\sin 30}{\tan x}=\frac{1}{4\cos 40\sin 20}$$$$\frac{1}{\tan x}=(\frac{1}{2\cos 40\sin 20}-\sqrt{3})$$$$\Rightarrow x=80°$$

Commented by I want to learn more last updated on 12/Oct/20

$$\mathrm{T}hankssir.Iappreciate$$

Commented by I want to learn more last updated on 12/Oct/20

$$\mathrm{sir},\mathrm{please}\mathrm{help}\mathrm{with}\mathrm{Q}117480$$

Answered by 1549442205PVT last updated on 12/Oct/20

Commented by 1549442205PVT last updated on 12/Oct/20

$$\mathrm{Construct}\mathrm{the}\mathrm{equilateral}\mathrm{triangle}$$$$\mathrm{BCD}\mathrm{such}\mathrm{that}\mathrm{D}\mathrm{lie}\mathrm{on}\mathrm{halfplane}$$$$\mathrm{with}\mathrm{the}\mathrm{edge}\mathrm{being}\mathrm{the}\mathrm{line}\mathrm{AC}{\mathrm{don}}^{\prime }\mathrm{t}$$$$\mathrm{containing}\mathrm{B}.\mathrm{Then}$$$$\widehat{\mathrm{BCD}}=\widehat{\mathrm{BDC}}=\widehat{\mathrm{CBD}}=60°.\mathrm{From}\mathrm{the}$$$$\mathrm{hypothesis}\widehat{\mathrm{BAC}}=\widehat{\mathrm{BCA}}=40°\mathrm{we}\mathrm{get}$$$$\widehat{\mathrm{ACD}}=20°,\widehat{\mathrm{ABD}}=100°-60°=40°$$$$\mathrm{BA}=\mathrm{BD}=\mathrm{BC}=\mathrm{CD}\Rightarrow \mathrm{\Delta }\mathrm{ABD}\mathrm{is}$$$$\mathrm{isosceles}\mathrm{at}\mathrm{B}\Rightarrow \widehat{\mathrm{BAD}}=\widehat{\mathrm{BDA}}=70°$$$$\Rightarrow \widehat{\mathrm{CAD}}=70°-40°=30°(\ast )$$$$\mathrm{Draw}\mathrm{the}\mathrm{ray}\mathrm{Cx}\mathrm{such}\mathrm{that}\widehat{\mathrm{BCx}}=30°$$$$\mathrm{M}=\left(\mathrm{AB}\right)\cap \mathrm{Cx}.\mathrm{Then}\widehat{\mathrm{AMC}}=180°-(40+70)$$$$=70°.\mathrm{Suppose}{\mathrm{O}}^{\prime }\mathrm{is}\mathrm{the}\mathrm{point}\mathrm{symetry}$$$$\mathrm{to}\mathrm{M}\mathrm{through}\mathrm{BC},\mathrm{E}=\mathrm{BC}\cap {\mathrm{MO}}^{\prime }\mathrm{then}$$$$\widehat{\mathrm{BME}}=\widehat{\mathrm{BMC}}-\widehat{\mathrm{CME}}=70-60=10°$$$$\Rightarrow \widehat{\mathrm{CBM}}=80°=\widehat{{\mathrm{CBO}}^{\prime }}(\mathrm{the}\mathrm{property}\mathrm{of}$$$$\mathrm{two}\mathrm{symetric}\mathrm{figures}).\mathrm{We}\mathrm{will}\mathrm{prove}$$$$\mathrm{that}\mathrm{O}\equiv {\mathrm{O}}^{\prime }.\mathrm{Indeed},\mathrm{we}\mathrm{have}\widehat{\mathrm{ABO}}=$$$$\widehat{\mathrm{ABC}}-\widehat{{\mathrm{CBO}}^{\prime }}=100-80=20°=\frac{1}{2}\widehat{\mathrm{ABD}}$$$${\mathrm{O}}^{\prime }\in \mathrm{CO}\mathrm{since}\widehat{\mathrm{BCO}}=\widehat{{\mathrm{BCO}}^{\prime }}=30°$$$$\widehat{{\mathrm{CO}}^{\prime }\mathrm{D}}=\widehat{{\mathrm{CO}}^{\prime }\mathrm{B}}=70°\mathrm{and}\widehat{{\mathrm{CDO}}^{\prime }}=\widehat{{\mathrm{CBO}}^{\prime }}=80°$$$$(\mathrm{since}{\mathrm{CO}}^{\prime }\mathrm{Dand}{\mathrm{CO}}^{\prime }\mathrm{B}\mathrm{are}\mathrm{two}\mathrm{symetric}$$$$\mathrm{figures}\mathrm{through}\mathrm{CO})\Rightarrow \widehat{{\mathrm{O}}^{\prime }\mathrm{DA}}=\widehat{\mathrm{ADC}}$$$$-\widehat{{\mathrm{CDO}}^{\prime }}=130-80=50°.\left(1\right)$$$$\mathrm{\Delta }{\mathrm{BO}}^{\prime }\mathrm{A}=\mathrm{\Delta }{\mathrm{BO}}^{\prime }\mathrm{D}(\mathrm{s}.\mathrm{a}.\mathrm{s})\Rightarrow \widehat{{\mathrm{BO}}^{\prime }\mathrm{A}}=\widehat{{\mathrm{BO}}^{\prime }\mathrm{D}}$$$$=140°\Rightarrow \widehat{{\mathrm{AO}}^{\prime }\mathrm{D}}=360-2.140=80°\left(2\right)$$$$\mathrm{From}\left(1\right)\left(2\right)\mathrm{we}\mathrm{get}\widehat{{\mathrm{O}}^{\prime }\mathrm{AD}}=180-(50+80)$$$$=50°\Rightarrow \widehat{{\mathrm{O}}^{\prime }\mathrm{AC}}=50-30=20°=\widehat{\mathrm{OAC}}\left(3\right)$$$$\mathrm{but}\widehat{{\mathrm{O}}^{\prime }\mathrm{CA}}=10°\left(4\right).\mathrm{From}\left(3\right)\left(4\right)\mathrm{we}\mathrm{infer}$$$${\mathrm{O}}^{\prime }\equiv \mathrm{O}.\mathrm{That}\mathrm{shows}\mathrm{that}\mathrm{x}=\widehat{\mathrm{CBO}}=80°$$

Commented by I want to learn more last updated on 12/Oct/20

$$Thankssir,iappreciate.$$

Commented by I want to learn more last updated on 12/Oct/20

$$\mathrm{sir},\mathrm{please}\mathrm{help}\mathrm{with}\mathrm{Q}117480$$

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