Question

ΔABC has three sides with these lengths: AB = 9, BC = 40, and CA = 41. What is the value of cos C?

Answer 1

Answer:

$cos(C) =\frac{40}{41}\\\\cos(C)=0.976$

Step-by-step explanation:

To find the $cos (C)$ we must use the cosine theorem.

The cosine theorem says that:

$c^2 = a^2 +b^2 -2abcos(C)$

In this case:

$c=AB = 9$

$b=CA = 41$

$a=BC = 40$

So

$9^2 = 40^2 +41^2 -2(40)*(41)cos(C)$

$9^2- 40^2- 41^2 =-2(40)*(41)cos(C)$

$cos(C) = -\frac{9^2- 40^2- 41^2}{2(40)*(41)}$

$cos(C) =\frac{40}{41}\\\\cos(C)=0.976$