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Question Number 95512 by bagjamath last updated on 25/May/20

$$\mathrm{If}\mathrm{in}\mathrm{a}△ABC,8R^2=a^2+b^2+c^2,\mathrm{then}$$$$\mathrm{the}△ABC\mathrm{is}$$

Commented by som(math1967) last updated on 25/May/20

$$\mathrm{Rt}\mathrm{angle}\mathrm{triangle}$$

Commented by som(math1967) last updated on 25/May/20

$${\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2={8\mathrm{R}}^2$$$${4\mathrm{R}}^2({\sin }^2\mathrm{A}+{\sin }^2\mathrm{B}+\sin {}^2\mathrm{c})={8\mathrm{R}}^2$$$${2\sin }^2\mathrm{A}+{2\sin }^2\mathrm{B}+{2\sin }^2\mathrm{C}=4$$$$\cos 2\mathrm{A}+\cos 2\mathrm{B}+{2\cos }^2\mathrm{C}=0$$$$2\cos (\mathrm{A}+\mathrm{B})\cos (\mathrm{A}-\mathrm{B})+{2\cos }^2\mathrm{C}=0$$$$-\mathrm{cosCcos}(\mathrm{A}-\mathrm{B})+{\cos }^2\mathrm{C}=0★$$$$-\mathrm{cosC}\{\cos (\mathrm{A}-\mathrm{B})-\cos (\mathrm{A}+\mathrm{B})=0★$$$$2\mathrm{cosCcos}\mathrm{Acos}\mathrm{B}=0$$$$\mathrm{eithercos}\mathrm{C}=0/\mathrm{cosB}=0/\mathrm{cosA}=0$$$$\therefore \mathrm{\angle }\mathrm{C}=90\mathrm{or}\mathrm{\angle }\mathrm{B}=90\mathrm{or}\mathrm{\angle }\mathrm{A}=90$$$$★\cos (\mathrm{A}+\mathrm{B})=\cos (\pi -\mathrm{C})=-\mathrm{cosC}$$

Commented by bagjamath last updated on 25/May/20

$$VeryNice$$$$ThankYouSir$$

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