Question

Solve for the missing length and the other two angles in the triangle below.

Answer 1

well you have b and c and you want to find side a right?

a^2 = b^2 + c^2 - 2bc*cos(A)

Answer 2

a^2=b^2+c^2-2*bc (Cos A)
a^2=7.4^2+15.8^2-2*7.4*15.8 (Cos(54))
a^2=166.95
a=12.92

Answer 3

Ugh I hope I did this right :/

Answer 4

awesome

then for part II you can use law of cosines to find angle C

Answer 5

okay hold on one sec

Answer 6

c²=a²+b²-2abCos(C)
15.8^2=12.92^2+7.4^2-2*12.92*7.4 (Cos(C))

C=6.28

Answer 7

I think :/

Answer 8

right setup, might be a calculator error

anyway I get angle C = 98.41 degrees

for part III since you have two angles you can just subtract 180 - the other two angles to find angle A

Answer 9

*to find angle B

Answer 10

180-98.41-54=27.59

Answer 11

well done

Answer 12

I multiply the x coordinates and y coordinates first right

Answer 13

yes

Answer 14

v1=-35

v2=24

Answer 15

uh, good, keep in mind these aren't the values of v1 and v2, but you're on the right track

to find v1 dot v2 you would just add -35 + 24

Answer 16

-11

Answer 17

good, so that's your dot product

next step is to find the magnitudes of v and w

magnitude = sqrt(x^2+y^2)

Answer 18

sqrt((-35)^2+24^2)

=sqrt1801

=42.44

Answer 19

v = (-7,8)
w = (5,3)

these are the x and y values you need to use

start with the magnitude of v

Answer 20

sqrt((-7)^2+8^2)
=sqrt113
=10.63

sqrt(5^2+3^2)
=sqrt34
=5.83

Answer 21

good but let's leave it in radical form to keep accuracy

now we just need to solve for cos(theta)

-11 = sqrt(113) * sqrt(34) * cos(theta)

solve for cos(theta) (not theta, just cos(theta))

Answer 22

this would be part 2 right

Answer 23

6.28

Answer 24

we're still on part I

Answer 25

okay

Answer 26

anyway, check your calculations again

-11 = sqrt(113) * sqrt(34) * cos(theta)

just divide both sides by sqrt(113) and sqrt(34) to get

-11/ [sqrt(113)*sqrt(34)] = ?

Answer 27

0.177

Answer 28

don't forget the - sign

-0.177 is the value of cos(theta)

then for part II simply take the arccos of that to get theta

Answer 29

100.2 degrees

Answer 30

good, that's it