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Question Number 36239 by Rio Mike last updated on 30/May/18

$$\mathrm{the}\mathrm{opposite}\mathrm{side}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{is}$$$$x+y,\mathrm{and}\mathrm{the}\mathrm{hypotenuse}\mathrm{is}3x+y$$$$\mathrm{Given}\mathrm{that}\mathrm{that}\mathrm{A}\mathrm{is}\mathrm{an}\mathrm{acute}\mathrm{angle}$$$$\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{Cos}\mathrm{A}$$

Commented by Rasheed.Sindhi last updated on 30/May/18

$$\mathrm{Opposite}\mathrm{side}\mathrm{to}\mathrm{\angle }\mathrm{A}=\mathrm{x}+\mathrm{y}$$$$\mathrm{Hypotenuse}=3\mathrm{x}+\mathrm{y}$$$$\mathrm{Adjacent}\mathrm{side}\mathrm{to}\mathrm{\angle }\mathrm{A}=\sqrt{{(3\mathrm{x}+\mathrm{y})}^2-{(\mathrm{x}+\mathrm{y})}^2}$$$$=\sqrt{{9\mathrm{x}}^2+6\mathrm{xy}+{\mathrm{y}}^2-{\mathrm{x}}^2-2\mathrm{xy}-{\mathrm{y}}^2}$$$$=\sqrt{{8\mathrm{x}}^2+4\mathrm{xy}}$$$$=2\sqrt{\mathrm{x}(2\mathrm{x}+\mathrm{y})}$$$$\cos \mathrm{A}=\frac{2\sqrt{\mathrm{x}(2\mathrm{x}+\mathrm{y})}}{3\mathrm{x}+\mathrm{y}}$$

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