In a triangle ABC , if cosA + 2cosB + cosC = 2 and the lengths of...

Given,

$AB = 7, BC = 3$.

$⇒ c = 7, a = 3$.

Apply cosine rule in the given equation $cosA + 2cosB + cosC = 2$, we get

$⇒ \left(\dfrac{{b}^{2} + {c}^{2} - {a}^{2}}{2bc}\right) + 2\left(\dfrac{{c}^{2} + {a}^{2} - {b}^{2}}{2ac}\right) + \left(\dfrac{{a}^{2} + {b}^{2} - {c}^{2}}{2ab}\right) = 2$

$⇒ \left(\dfrac{{b}^{2} + 49 - 9}{14b}\right) + \left(\dfrac{49 + 9 - {b}^{2}}{21}\right) + \left(\dfrac{9 + {b}^{2} - 49}{6b}\right) = 2$

$⇒ \left(\dfrac{{b}^{2} + 40}{14b}\right) + \left(\dfrac{58 - {b}^{2}}{21}\right) + \left(\dfrac{{b}^{2} - 40}{6b}\right) = 2$

$⇒ {b}^{3} - 5{b}^{2} - 16b + 80 = 0$

$⇒ \left(b - 4\right)\left(b + 4\right)\left(b - 5\right) = 0$

$⇒ b = - 4, 4, 5$.

But $b \ne - 4$.

Also, if $b = 4$ triangle cannot be constructed.

That means $b = 5$.

$⇒ cosA - cosC = \left(\dfrac{{b}^{2} + {c}^{2} - {a}^{2}}{2bc}\right) - \left(\dfrac{{a}^{2} + {b}^{2} - {c}^{2}}{2ab}\right)$

$⇒ cosA - cosC = \left(\dfrac{{5}^{2} + {7}^{2} - {3}^{2}}{2\left(5 \times 7\right)}\right) - \left(\dfrac{{3}^{2} + {5}^{2} - {7}^{2}}{2\left(3 \times 5\right)}\right)$

$⇒ cosA - cosC = \left(\dfrac{65}{70}\right) - \left(\dfrac{- 15}{30}\right) = \dfrac{195 + 105}{210} = \dfrac{300}{210}$

$⇒ cosA - cosC = \dfrac{10}{7}$