Question

I still need help on one more please

Answer 1

I got 8.49 for the missing side but since one side is sqrt. im confused @jim_thompson5910

Answer 2

the missing side should be a whole number

Answer 3

hint: we have a 30-60-90 triangle

Answer 4

so then just 8?

Answer 5

not quite

Answer 6

4?

Answer 7

let me know if this helps or not

http://hotmath.com/hotmath_help/topics/30-60-90-triangles/triangle1.gif

Answer 8

a^2 + b^2=c^2 because it is a right angle.
So just do (6)^2+6rad(3)= and square root the answer.

Answer 9

i got 6

Answer 10

and that really did help clear things up

Answer 11

didnt**

Answer 12

you should get 12

Answer 13

or you can use the pythagorean theorem

a^2 + b^2 = c^2
6^2 + (6*sqrt(3))^2 = c^2
36 + 36*3 = c^2
36 + 108 = c^2
144 = c^2
c^2 = 144
c = sqrt(144)
c = 12

Answer 14

so again, the missing side is 12

Answer 15

okay

Answer 16

so,

cos(B) = adj/hyp
cos(B) = 6/12
cos(B) = 1/2

Answer 17

2.44 as my answer

Answer 18

idk how you got that

Answer 19

me neither

Answer 20

do you see how I got 1/2 ?

Answer 21

yeah

Answer 22

one side is \(6\) the other sides is \(6\sqrt3\)
that makes the hypotenuse \(2\times 6=12\) since the ratios of such a right triangle (30-60-90\) are \[1:\sqrt3:2\] meaning the hypotenuse is twice the short leg

Answer 23

Sqrt.3/2 would be the right answer then?

Answer 24

what are you trying to find, sine or cosine

Answer 25

cos

Answer 26

the sine is \(\frac{\sqrt3}{2}\) as in "opposite over hypotenuse"

Answer 27

the cosine is \(\frac{1}{2}\)
oh hold on, let me check which angle it is

Answer 28

yeah angle B, cosine is \(\frac{1}{2}\)

Answer 29

okay thank you(: