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If cos a + 2cos b + cos c = 2 then a, b, c are |
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Answer» => cos A + cos C = 2 (1 - cos B) => (b²+c²-a²)/2bc + (c²+a²-b²)/2ca = 2[1 - (a²+b²-c²)/2ab] => a(b²+c²-a²) + b(c²+a²-b²) = 2c[2ab - (a²+b²-c²)] => ab² + ac² -a³ + bc² + ba² - b³ = 2c[2ab - (a²+b²-c²)] => ab² + ba² + ac² + bc² - a³ - b³ = 2c[2ab - (a²+b²-c²)] => ab(a+b) + c²(a+b) - (a+b)(a²-ab+b²) = 2c[2ab - (a²+b²-c²)] => (a+b)(ab+c² - a²+ab-b²) = 2c[2ab - (a²+b²-c²)] => (a+b)[2ab - (a²+b²-c²)] = 2c[2ab - (a²+b²-c²)] => a+b = 2c => SIDES of the triangle are in A.P. |
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