Question

In a ∆ABC, if a = 3, b = 5, c = 7, find cosA, cosB and cosC.

Answer 1

cos A=
$\frac{3}{7}$
cos B=
$\frac{5}{7}$
cosC=
$\frac{5}{7}$

Answer 2

In a Triangle ABC,

If AB = c, BC = a, AC = b

By cosine rule,

c² = a² + b² - 2abcosC

2abcosC= a² + b² - c²

$cosC = \frac{ {a}^{2} + {b}^{2} - {c}^{2} }{2ab} \\ \\ cos B = \frac{ {c}^{2} + {a}^{2} - {b}^{2} }{2ac} \\ \\ cosA = \frac{ {c}^{2} + {b}^{2} - {a}^{2} }{2bc}$

Given

a = 3

b = 5

c = 7

Now,

$cosC = \frac{ {a}^{2} + {b}^{2} - {c}^{2} }{2ab} \\ \\ cosC = \frac{ {3}^{2} + {5}^{2} - {7}^{2} }{2(5)(3)} \\ \\ cosC = \frac{25+ 9 - 49}{30} \\ \\ cosC = \frac{ - 15}{30} \\ \\ cosC = \frac{ - 1}{2}$

Therefore, cosC = - 1/2

$cos B = \frac{ {c}^{2} + {a}^{2} - {b}^{2} }{2ac} \\ \\ cos B = \frac{ {7}^{2} + {3}^{2} - {5}^{2} }{2(7)(3)} \\ \\ cos B = \frac{49 + 9 - 25}{42} \\ \\ cos B = \frac{33}{42} = \frac{11}{14}$

Therefore, cosB = 11/14

$cosA = \frac{ {c}^{2} + {b}^{2} - {a}^{2} }{2bc} \\ \\ cosA= \frac{ {7}^{2} + {5}^{2} - {3}^{2} }{2(7)(5)} \\ \\ cosA= \frac{49 + 25 - 9}{70} = \frac{65}{70} = \frac{13}{14}$

Therefore, cosA = 13/14