Question

Triangle ABC hac three sides with these lengths: AB=9, BC=40, and CA=41. What is the value of cos c

Answer 1

Answer:

$cos c = \frac{40}{41}$

Step-by-step explanation:

We have a (right-angled) triangle ABC with the following side lengths:

AB=9,

BC=40; and

CA=41.

We are to find the value of cos c. We know that $cos$∅$= \frac{base}{hypotenuse}$.

In this case, since we have to find the angle c so the AB will be the opposite, BC the base and AC will be the hypotenuse.

$cos c = \frac{BC}{AC}$

Therefore, $cos c = \frac{40}{41}$

Answer 2

Answer

cos c = 40/41

Step by step explanation

It is a right triangle.

AB^2 + BC^2 = AC^2

9^2 + 40^2 = 41^2

81 + 1600 = 1681

1681 = 1681

Cos C = Opposite/Hypotenuse

Cos C = 40/41             [opposite = 40 and Hypotenuse = 41]

Thank you.