In a triangle ABC, if cos A + 2 cos B + cos C = 2 and the lengths of the sides opposite to angles A and C are 3 and 7 respectively, then cos A - cos C is equal to [2023]

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Solution

Q.
In a triangle ABC, if $\cos A + 2 \cos B + \cos C = 2$ and the lengths of the sides opposite to angles A and C are 3 and 7 respectively, then $\cos A - \cos C$ is equal to [2023]

1 $\frac{5}{7}$

2 $\frac{10}{7}$

3 $\frac{3}{7}$

4 $\frac{9}{7}$

Ans.
(2)

$We have, \cos A + 2 \cos B + \cos C = 2$

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

$\Rightarrow [\frac{b^{2} + 49 - 9}{14 b}] + 2 [\frac{49 + 9 - b^{2}}{42}] + [\frac{9 + b^{2} - 49}{6 b}] = 2$ $[∵ a = 3, c = 7]$

$\Rightarrow \frac{b^{2} + 40}{14 b} + \frac{58 - b^{2}}{21} + \frac{- 40 + b^{2}}{6 b} = 2$

$\Rightarrow b^{3} - 5 b^{2} - 16 b + 80 = 0$

$\Rightarrow (b - 4) (b + 4) (b - 5) = 0$

$\Rightarrow b = - 4$ or $4$ or $5$

$b \neq - 4$ $(∵ b cannot be - 4)$

$For b = 4, triangle cannot be constructed$

$∴$ $b = 5$

$Now, \cos A - \cos C = \frac{b^{2} + c^{2} - a^{2}}{2 b c} - \frac{a^{2} + b^{2} - c^{2}}{2 a b}$

$= \frac{49 + 25 - 9}{2 \times 7 \times 5} - \frac{9 + 25 - 49}{2 \times 3 \times 5} = \frac{13}{14} + \frac{1}{2} = \frac{10}{7}$

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