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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

$$let\mathrm{\angle }DBM=\mathrm{\angle }MBC=\varphi$$$$\mathrm{\angle }ECM=\mathrm{\angle }MCB=\theta$$$$2\theta +2\varphi =90°$$$$\Rightarrow \theta =45°-\varphi$$$$\Rightarrow \mathrm{\angle }BMC=180°-45°=135°$$$$\frac{6}{sin\varphi }=\frac{2\sqrt{2}}{sin(45°-\varphi )}=\frac{x}{sin135°}$$$$\sqrt{2}sin\varphi =3(sin45°cos\varphi -cos45°sin\varphi )$$$$sin\varphi =\frac{3}{2}(cos\varphi -sin\varphi )$$$$5sin\varphi =3cos\varphi$$$$letsin\varphi =t,t>0$$$$5t=3\sqrt{1-t^2}$$$$25t^2=9(1-t^2)$$$$34t^2=9$$$$t=\frac{3}{\sqrt{34}}=sin\varphi$$$$\frac{6}{\frac{3}{\sqrt{34}}}=\frac{x}{\frac{1}{\sqrt{2}}}$$$$x=BC=2\sqrt{17}$$

Commented by TawaTawa last updated on 22/Aug/19

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

Commented by TawaTawa last updated on 23/Aug/19

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

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

$$2\theta +2\varphi =90°$$$$\Rightarrow \theta +\varphi =45°$$$$\mathrm{\angle }BMC=180°-(\theta +\varphi )=180°-45°=135°$$

Commented by TawaTawa last updated on 23/Aug/19

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

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