Question

Help!!

Answer 1

Law of cosines
\[c^{2} = a^{2}+b^{2}-2ab \cos \theta\]
where theta is always included angle opposite of side c

Answer 2

yea but i m getting a negative answer..

Answer 3

like...\[c ^{2}=(3)^{2}+(2\sqrt{2})^{2}-2(3)(2\sqrt{2})(3\lambda/4)\]

Answer 4

c^2=9+8-39.98
c=N/A

Answer 5

check your cos(theta)....if theta > 90 then it will be neg

Answer 6

ohh okay..

Answer 7

so its
c^2=9+8+39.98
c^2=56.98
c=7.548 <-- correct?

Answer 8

no the 39.98 is way too big

Answer 9

remember c< a+b
--> c < 5.83

Answer 10

hmmm...

Answer 11

did you take cos(3pi/4) ?
looks like you just plugged in 3pi/4

Answer 12

i got that answer..!! :)

Answer 13

so its like \[c ^{2}=3^{2}+2\sqrt{2})^{2}-2(3)(2\sqrt{2})(-\sqrt{2}/2)\]

Answer 14

c^2=9+8+12
c=5.385165

Answer 15

that looks better