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Question Number 9370 by tawakalitu last updated on 03/Dec/16

Answered by geovane10math last updated on 03/Dec/16

$$\mathrm{There}\mathrm{are}\mathrm{infinite}\mathrm{triangles}\mathrm{with}\mathrm{sizes}10$$$$\mathrm{and}15.\mathrm{It}\mathrm{depends}\mathrm{of}\mathrm{inclination}\mathrm{of}X\widehat{ZM}.$$$$\mathrm{I}\mathrm{think}\mathrm{that}\mathrm{the}\mathrm{value}\mathrm{of}w\mathrm{depends}\mathrm{of}$$$$X\widehat{ZM}.$$$$X\widehat{ZM}=\alpha$$$${\mathrm{Cosine}}^{\prime }\mathrm{s}\mathrm{Law}:$$$${\mathit{w}}^2={15}^2+{10}^2-2\cdot 10\cdot 15\cdot \cos \mathit{\alpha }$$$${\mathit{w}}^2=225+100-300\cdot \cos \mathit{\alpha }$$$$$$$$\mathit{w}=\sqrt{325-300\cdot \cos \mathit{\alpha }}$$

Commented by geovane10math last updated on 03/Dec/16

$$DependsofXM=8.$$

Commented by tawakalitu last updated on 03/Dec/16

$$\mathrm{I}\mathrm{appreciate}\mathrm{your}\mathrm{effort}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}.$$

Answered by mrW last updated on 03/Dec/16

$$\mathrm{\Delta }\mathrm{YXM}\mathrm{and}\mathrm{\Delta }\mathrm{YZX}\mathrm{are}\mathrm{similar},$$$$\mathrm{because}\mathrm{they}\mathrm{have}\mathrm{two}\mathrm{equal}\mathrm{angles}.$$$$\frac{\mid \mathrm{YX}\mid }{\mid \mathrm{YZ}\mid }=\frac{\mid \mathrm{XM}\mid }{\mid \mathrm{ZX}\mid }=\frac{\mid \mathrm{MY}\mid }{\mid \mathrm{XY}\mid }$$$$\frac{\mathrm{w}}{15}=\frac{8}{10}=\frac{\mid \mathrm{MY}\mid }{\mathrm{w}}$$$$\mathrm{w}=\frac{8}{10}\times 15=12\mathrm{cm}$$$$\mid \mathrm{MY}\mid =\frac{8}{10}\times 12=\frac{48}{5}=9.6\mathrm{cm}$$$$\mid \mathrm{MZ}\mid =15-\mid \mathrm{MY}\mid =15-9.6=5.4\mathrm{cm}$$

Commented by tawakalitu last updated on 03/Dec/16

$$\mathrm{Thanks}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}.$$

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