Question

Solve the triangle. Round lengths to the nearest tenth and angle measures to the nearest degree.
a=7
b=8
c=7

Answer 1

use cosine laws...

Answer 2

ok, so far i know to use c^2=a^2+b^2-2abcosC. can u see if i'm doing this right?

Answer 3

yes y not..show your work..

Answer 4

ok, I have 7^2=4^2+8^2-2*4*8cosC => 49-16-64=64cosC

Answer 5

use correct values..you plug in a=4
but it is a=7

Answer 6

oh ok, so if a=7, c=4?

Answer 7

still you plug in wrong no if a=7 and c=4

but it is true if a=4 andc =7

Answer 8

oh my bad, i meant to put a=4, b=8, and c=7

Answer 9

so 7^2=4^2+8^2-2*4*8cosC => 49-16-64= 64cosC

Answer 10

xactly right plugging this time..

Answer 11

thanks. and to find 64cosC, do i replace it with 180?

Answer 12

no...now simplyfy the left side and then divide by 64 to free cosC..then take cos^-1

Answer 13

ok, so 49-16-64=-31, (-31/64)= -.48437(or -0.5) , cos^-1(-0.5)=120

Answer 14

i got it now. I actually gotta use the formula b^2=a^2+c^2-2accosB. so i have 8^2=4^2+7^2-2*4*7cosB => -1/-56= cos^-1(.0178)=88.9=89 degrees.

Answer 15

thank u. i think i can do the rest of it on my own now.

Answer 16

all i can do is use the sine laws to solve the rest.

Answer 17

okay..its allright..