Question

Can you use the Law of cosines in the triangle below? Why or why not?

Answer 1

@wolf1728

Answer 2

well, what do you think?

Answer 3

I don't know.. that's why I asked this question

Answer 4

do you know the law of cosines?

Answer 5

no

Answer 6

heheh, it helps to know it
you can't really find what you don't know what it's

Answer 7

yeah.. I really don't get this lesson

Answer 8

http://mathworksheetsgo.com/sheets/trigonometry/advanced/law-of-sines-and-cosines/images/law-of-cosines-picture-formula.jpg

Answer 9

so there is 3?

Answer 10

well, sorta, not really is one, applied to each angle

what it really says in the notation is
one side that's facing off an angle, SQUARED
equals
the sum of the other 2 sides SQUARED minus ( 2 times the product of the same 2 sides times the cosine of the angle facing it off)

Answer 11

so if you want to find side "a", the other 2 sides are "b and c" and the cos(A)
if you want to find "b", then the other 2 sides will be "a and c" and the cos(B)
and so on

Answer 12

ok so how do I know if I can use it for the triangle or not?

Answer 13

your triangle has 3 sides given
no angles provided

let's say

let's try to find angle B
just by knowing all 3 sides

do you think you can factor and simplify the Law of Cosines to find angle B?

Answer 14

yeah you just plug it in to the formula right?

Answer 15

well, lemme write it for you in terms of side "b"

Answer 16

\(\bf b^2 = a^2+c^2-2ac\ cos(B) \implies b = \sqrt{a^2+c^2-2ac\ cos(B)}\\
21 = \sqrt{22^2+15^2-(2(22)(15)\ cos(B)}\)

Answer 17

what do you think? can we find angle B?

Answer 18

yeah

Answer 19

or \(\bf 21^2 = 22^2+15^2-(2(22)(15)\ cos(B)\)

Answer 20

ok, so, let's see what we get for angle B then :)

Answer 21

66

Answer 22

well, I got a different number :), lemme rewrite it some

Answer 23

sorry Im horrible at math

Answer 24

ohhh my bad

Answer 25

as usual, I missed the negative denominator, yes you're correct, is 66 degrees :)

Answer 26

so as you can see, yes, we can use it in this triangle

Answer 27

oh ok thanks!! but what do I put for why?

Answer 28

like for the why or why not

Answer 29

we can use it because the Law of cosines can be refactored and simplified to
\(\bf b^2 = a^2+c^2-2ac\ cos(B) \implies \cfrac{b^2-a^2-c^2}{-2ac} = cos(B)\\
cos^{-1}(cos(B)) = cos^{-1}\left(\cfrac{b^2-a^2-c^2}{-2ac}\right) =
\measuredangle B\)

to get any angle once we have all sides

Answer 30

ok thank you!

Answer 31

yw