In-ABC-a-cos-A-b-cos-B-c-cos-C-then-ABC-is-A-irregular-sides-acute-angled-triangle-B-obtuse-angled-triangle-C-right-angled-triangle-D-equilateral-triangle-E-isoceles-triangle-

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Question Number 117500 by ZiYangLee last updated on 12/Oct/20

$$\mathrm{In}\mathrm{\Delta }\mathrm{ABC},\frac{a}{\cos A}=\frac{b}{\cos B}=\frac{c}{\cos C},$$$$\mathrm{then}\mathrm{\Delta }\mathrm{ABC}\mathrm{is}$$$$A.\mathrm{irregular}\mathrm{sides}\mathrm{acute}-\mathrm{angled}\mathrm{triangle}$$$$B.\mathrm{obtuse}-\mathrm{angled}\mathrm{triangle}$$$$C.\mathrm{right}-\mathrm{angled}\mathrm{triangle}$$$$D.\mathrm{equilateral}\mathrm{triangle}$$$$E.\mathrm{isoceles}\mathrm{triangle}$$

Answered by som(math1967) last updated on 12/Oct/20

$$\mathrm{D}\mathrm{equilateral}.\mathrm{ans}$$$$\frac{\mathrm{a}}{\mathrm{cosA}}=\frac{\mathrm{b}}{\mathrm{cosB}}=\frac{\mathrm{c}}{\mathrm{cosC}}$$$$\frac{2\mathrm{abc}}{{\mathrm{b}}^2+{\mathrm{c}}^2-{\mathrm{a}}^2}=\frac{2\mathrm{abc}}{{\mathrm{c}}^2+{\mathrm{a}}^2-{\mathrm{b}}^2}=\frac{2\mathrm{abc}}{{\mathrm{a}}^2+{\mathrm{b}}^2-{\mathrm{c}}^2}$$$${\mathrm{b}}^2+{\mathrm{c}}^2-{\mathrm{a}}^2={\mathrm{c}}^2+{\mathrm{a}}^2-{\mathrm{b}}^2={\mathrm{a}}^2+{\mathrm{b}}^2-{\mathrm{c}}^2$$$$\because {\mathrm{b}}^2+{\mathrm{c}}^2-{\mathrm{a}}^2={\mathrm{c}}^2+{\mathrm{a}}^2-{\mathrm{b}}^2$$$${2\mathrm{b}}^2={2\mathrm{a}}^2\Rightarrow \mathrm{b}=\mathrm{a}$$$$\mathrm{again}{\mathrm{c}}^2+{\mathrm{a}}^2-{\mathrm{b}}^2={\mathrm{a}}^2+{\mathrm{b}}^2-{\mathrm{c}}^2$$$$\therefore {2\mathrm{c}}^2={2\mathrm{b}}^2\Rightarrow \mathrm{c}=\mathrm{b}$$$$\therefore \mathrm{a}=\mathrm{b}=\mathrm{c}$$

Commented by ZiYangLee last updated on 12/Oct/20

$$\mathrm{Omg}\mathrm{so}\mathrm{easy}!$$

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