SATHEE: Properties of Triangles 2 Question 14

Properties of Triangles 2 Question 14

14. For a $△ A B C$ , it is given that $\cos A + \cos B + \cos C = \frac{3}{2}$ . Prove that the triangle is equilateral.

(1984, 4M)

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Solution:

Let $a, b, c$ are the sides of a $△ A B C$ .

Given, $\cos A + \cos B + \cos C = \frac{3}{2}$

$\Rightarrow \frac{b^{2} + c^{2} - a^{2}}{2 b c} + \frac{a^{2} + c^{2} - b^{2}}{2 a c} + \frac{a^{2} + b^{2} - c^{2}}{2 a b} = \frac{3}{2}$

$\Rightarrow a b^{2} + a c^{2} - a^{3} + b a^{2} + b c^{2} - b^{3}$

$+ c a^{2} + c b^{2} - c^{3} = 3 a b c$

$\Rightarrow a (b - c)^{2} + b (c - a)^{2} + c (a - b)^{2}$

$= \frac{(a + b + c)}{2} [(a - b)^{2} + (b - c)^{2} + (c - a)^{2}]$

$\Rightarrow (a + b - c) (a - b)^{2} + (b + c - a) (b - c)^{2}$

$+ (c + a - b) (c - a)^{2} = 0$

[as we know, $a + b - c > 0, b + c - a > 0, c + a - b > 0$ ] $∴$ Each term on the left of equation has positive coefficient multiplied by perfect square, each term must be separately zero.

$\Rightarrow$

$a = b = c$

$∴$ Triangle is an equilateral.

Properties of Triangles 2 Question 14

14. For a $△ A B C$ , it is given that $\cos A + \cos B + \cos C = \frac{3}{2}$ . Prove that the triangle is equilateral.

Solution:

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Properties of Triangles 2 Question 14

14. For a △ABC, it is given that cos⁡A+cos⁡B+cos⁡C=32. Prove that the triangle is equilateral.

Solution:

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14. For a $△ A B C$ , it is given that $\cos A + \cos B + \cos C = \frac{3}{2}$ . Prove that the triangle is equilateral.