Question

I'm trying to find "a" using the law of cosines when I have A (50 degrees), b= 13, and c = 6. The equation that they've prompted me to use is a^2 = b^2 + c^2 -2(b)(c) (cosA). The final answer I keep getting is somewhere around 26-27, but the available choices are not within that range. What am I doing incorrectly?

Answer 1

is your calculator mode in degrees or radians?

Answer 2

Degree.

Answer 3

your calculations are in error then, im gettig something closer to 10

Answer 4

Is there a certain order I would go about solving this then?

Answer 5

not really ... show me yuor work

Answer 6

Alright, hold on. I need to rewrite it so it's legible.

Answer 7

typing is legible :)

Answer 8

Alright @amistre64 sorry... Here's a link to the work. http://prntscr.com/6xat6q

Answer 9

can you explain your line 3?

Answer 10

yeah, your line 3 is just making stuff up ... you stopped doing math

Answer 11

13^2 + 6^2 -2(13)(6) K

is NOT

[13^2 + 6^2 -2(13)(6)] K

Answer 12

+ Line three was created through combining like terms. That is, 169 + 36 - 26.

Answer 13

no it wasnt ...

Answer 14

it was fabricated using imagination instead of proper math.

Answer 15

If you insist.

Answer 16

when does:

a + bx = (a+b)x ?

Answer 17

I see.

Answer 18

do: 13^2 + 6^2

then subtract 2(13)(6) cos(50)

then take the sqrt

Answer 19

Would the 6 be by itself?

Answer 20

no, 2(13)(6) cos(50)

is all one term, it is a product of 4 factors.

Answer 21

the order in which we multiply it out is immaterial

13 cos(50) (2)(6)

or whatever

Answer 22

13^2 = 169
6^2 = 36
2(13)(6) cos(50) = 156 cos(50) = 100.2748....

Answer 23

169 + 36 - 100.2748... = 104.7252....

c = sqrt(104.7252....) ... which is close to 10

Answer 24

I see. This makes a lot more sense, thank you.

Answer 25

a = sqrt(104...) typoed the letter :) good luck