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Question Number 67033 by TawaTawa last updated on 22/Aug/19

Commented by Tony Lin last updated on 22/Aug/19

let∠DBM=∠MBC=φ ∠ECM=∠MCB=θ 2θ+2φ=90° ⇒θ=45°−φ ⇒∠BMC=180°−45°=135° (6/(sinφ))=((2(√2))/(sin(45°−φ)))=(x/(sin135°)) (√2)sinφ=3(sin45°cosφ−cos45°sinφ) sinφ=(3/2)(cosφ−sinφ) 5sinφ=3cosφ let sinφ=t, t>0 5t=3(√(1−t^2 )) 25t^2 =9(1−t^2 ) 34t^2 =9 t=(3/(√(34)))=sinφ (6/(3/(√(34))))=(x/(1/(√2))) x=BC=2(√(17))

$${let}\angle{DBM}=\angle{MBC}=\phi \\ $$$$\:\:\:\:\:\angle{ECM}=\angle{MCB}=\theta \\ $$$$\mathrm{2}\theta+\mathrm{2}\phi=\mathrm{90}° \\ $$$$\Rightarrow\theta=\mathrm{45}°−\phi \\ $$$$\Rightarrow\angle{BMC}=\mathrm{180}°−\mathrm{45}°=\mathrm{135}° \\ $$$$\frac{\mathrm{6}}{{sin}\phi}=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{{sin}\left(\mathrm{45}°−\phi\right)}=\frac{{x}}{{sin}\mathrm{135}°} \\ $$$$\sqrt{\mathrm{2}}{sin}\phi=\mathrm{3}\left({sin}\mathrm{45}°{cos}\phi−{cos}\mathrm{45}°{sin}\phi\right) \\ $$$${sin}\phi=\frac{\mathrm{3}}{\mathrm{2}}\left({cos}\phi−{sin}\phi\right) \\ $$$$\mathrm{5}{sin}\phi=\mathrm{3}{cos}\phi \\ $$$${let}\:{sin}\phi={t},\:{t}>\mathrm{0} \\ $$$$\mathrm{5}{t}=\mathrm{3}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} } \\ $$$$\mathrm{25}{t}^{\mathrm{2}} =\mathrm{9}\left(\mathrm{1}−{t}^{\mathrm{2}} \right) \\ $$$$\mathrm{34}{t}^{\mathrm{2}} =\mathrm{9} \\ $$$${t}=\frac{\mathrm{3}}{\sqrt{\mathrm{34}}}={sin}\phi \\ $$$$\frac{\mathrm{6}}{\frac{\mathrm{3}}{\sqrt{\mathrm{34}}}}=\frac{{x}}{\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}} \\ $$$${x}={BC}=\mathrm{2}\sqrt{\mathrm{17}} \\ $$

Commented by TawaTawa last updated on 22/Aug/19

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\: \\ $$

Commented by TawaTawa last updated on 23/Aug/19

Sir, looking at the solution^� How do i know angle BMC = 180 − 45. Thanks sir.

$$\mathrm{Sir},\:\mathrm{looking}\:\mathrm{at}\:\mathrm{the}\:\mathrm{solutio}\bar {\mathrm{n}}\:\: \\ $$$$\mathrm{How}\:\mathrm{do}\:\mathrm{i}\:\mathrm{know}\:\mathrm{angle}\:\:\:\:\mathrm{BMC}\:=\:\mathrm{180}\:−\:\mathrm{45}. \\ $$$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$

Commented by Tony Lin last updated on 23/Aug/19

2θ+2φ=90° ⇒θ+φ=45° ∠BMC=180°−(θ+φ)=180°−45°=135°

$$\mathrm{2}\theta+\mathrm{2}\phi=\mathrm{90}° \\ $$$$\Rightarrow\theta+\phi=\mathrm{45}° \\ $$$$\angle{BMC}=\mathrm{180}°−\left(\theta+\phi\right)=\mathrm{180}°−\mathrm{45}°=\mathrm{135}° \\ $$

Commented by TawaTawa last updated on 23/Aug/19

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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