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Question Number 148118 by mathdanisur last updated on 25/Jul/21

Answered by liberty last updated on 25/Jul/21

Commented by liberty last updated on 25/Jul/21

$$AM=\sqrt{{12}^2-4^2}=\sqrt{16\times 8}=8\sqrt{2}$$$$\iff \cos \theta =\frac{4}{12}=\frac{7}{4+x}$$$$\frac{1}{3}=\frac{7}{4+x}\Rightarrow 4+x=21$$$$\Rightarrow x=17=BM$$

Commented by mathdanisur last updated on 25/Jul/21

$$thankyouSer$$

Answered by Olaf_Thorendsen last updated on 25/Jul/21

$${\mathrm{AM}}^2={\mathrm{AC}}^2-{\mathrm{MC}}^2$$$${\mathrm{AM}}^2={(5+7)}^2-4^2=144-16=128$$$$\mathrm{AM}=8\sqrt{2}$$$$$$$$\tan \stackrel{\wedge }{\mathrm{C}}=\frac{\mathrm{AM}}{\mathrm{MC}}=\frac{\mathrm{BN}}{\mathrm{NC}}$$$$\frac{8\sqrt{2}}{4}=\frac{\mathrm{BN}}{7}\Rightarrow \mathrm{BN}=14\sqrt{2}$$$$$$$$\tan \stackrel{\wedge }{\mathrm{A}}=\frac{\mathrm{BN}}{\mathrm{AN}}=\frac{14\sqrt{2}}{5}$$$$\tan \left(\mathrm{MAC}\right)=\frac{\mathrm{MC}}{\mathrm{AM}}=\frac{4}{8\sqrt{2}}=\frac{1}{2\sqrt{2}}$$$$$$$$\tan \left(\mathrm{BAM}\right)=\tan (\stackrel{\wedge }{\mathrm{A}}-\mathrm{MAC})$$$$\tan \left(\mathrm{BAM}\right)=\frac{\tan \stackrel{\wedge }{\mathrm{A}}-\tan \left(\mathrm{MAC}\right)}{1+\tan \stackrel{\wedge }{\mathrm{A}}.\tan \left(\mathrm{MAC}\right)}$$$$\tan \left(\mathrm{BAM}\right)=\frac{\frac{14\sqrt{2}}{5}-\frac{1}{2\sqrt{2}}}{1+\frac{14\sqrt{2}}{5}.\frac{1}{2\sqrt{2}}}=\frac{17}{8\sqrt{2}}$$$$\tan \left(\mathrm{BAM}\right)=\frac{\mathrm{BM}}{\mathrm{AM}}$$$$\frac{17}{8\sqrt{2}}=\frac{\mathrm{BM}}{8\sqrt{2}}\Rightarrow \mathrm{BM}=17$$

Commented by mathdanisur last updated on 25/Jul/21

$$ThankyouSer$$

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