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

In ΔABC, (a/(cos A))=(b/(cos B))=(c/(cos C)),  then ΔABC is   A. irregular sides acute-angled triangle  B. obtuse-angled triangle  C. right-angled triangle  D. equilateral triangle  E. isoceles triangle

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

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

D equilateral.ans  (a/(cosA))=(b/(cosB))=(c/(cosC))  ((2abc)/(b^2 +c^2 −a^2 ))=((2abc)/(c^2 +a^2 −b^2 ))=((2abc)/(a^2 +b^2 −c^2 ))  b^2 +c^2 −a^2 =c^2 +a^2 −b^2 =a^2 +b^2 −c^2   ∵b^2 +c^2 −a^2 =c^2 +a^2 −b^2   2b^2 =2a^2 ⇒b=a  again c^2 +a^2 −b^2 =a^2 +b^2 −c^2   ∴2c^2 =2b^2 ⇒c=b  ∴a=b=c

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

Commented by ZiYangLee last updated on 12/Oct/20

Omg so easy!

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

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