Question

True?

Answer 1

I got something a little different than 22.5, are you familiar with the law of cosines?

Answer 2

if you let theta be capital C, and lowercase c the side across from C (so c = 5.3) while a and b are the other two sides (4 and 7) you should be able to plug everything into the formula and solve for capital C (theta)

Answer 3

5.3^2=4^2+7^2-2(4*7)cosC
C=0.85
C=5.43

Answer 4

\(C=cos^{-1}(\frac{a^2+b^2-c^2}{2ab})\)

Answer 5

better to convert your solutions to degrees since that's what's given in the problem

Answer 6

C=0.85 degrees
C=5.43 degrees

Answer 7

hmm,I got a diff answer

Answer 8

0.85 is in radians, you need to convert this to degrees

Answer 9

Ah I c...makes sense,now

Answer 10

48.70
311

Answer 11

so false

Answer 12

good, 48.70 degrees so the original answer is false

Answer 13

B

Answer 14

good

Answer 15

This would be the same concept as the first question right

Answer 16

yes, just be sure you are solving for cos(theta) not just theta

Answer 17

10 would be c right

Answer 18

10^2=8^2+7^2-2*7*8 Cos

Answer 19

I got something a little different, remember that c^2 is the side across from the angle

so

7^2 = 10^2+8^2 - 2(10)(8)cos(x)

Answer 20

0.77
5.51

Answer 21

keep in mind they are not asking for the value for x, they asking for cos(x) so you need to solve for cos(x)

Answer 22

B

Answer 23

0.72

Answer 24

good

Answer 25

so solving for c^2 this time right

Answer 26

c^2=2^2+3^2-(2*2*3)cos(pi/3)

Answer 27

im getting eroor on the calc though

Answer 28

not sure if i did something rong

Answer 29

you there?

Answer 30

yeah i'm still there i was just working on someone else's problem ;_;

anyway we can try doing some manual work if the calculator is being a butt

c^2=2^2+3^2-(2*2*3)cos(pi/3)

becomes

c^2 = 4 + 9 - (12)(1/2)

so c^2 = 13 - 6

c = ?

Answer 31

B

Answer 32

excellent

Answer 33

True

Answer 34

good

Answer 35

B

Answer 36

hm. I got something a little different

if theta (the angle across from b) is obtuse, then it must be at least 90 right

try applying the law of cosines

b^2 > a^2 + c^2 - 2ac * cos(90)

lmk what you get when you simplify this

Answer 37

I got D

Answer 38

awesome that's what i got too

Answer 39

idk if its a or c :/

Answer 40

well when you plug in 90 for the right angle we get

c^2 = a^2 + b^2 - 2ac(cos*90))

look like any familiar formula to you?

Answer 41

pyth?

Answer 42

A

Answer 43

awesome so A

Answer 44

C

Answer 45

awesome

Answer 46

B and C

Answer 47

keep in mind pythagorean states that

c^2 = a^2 + b^2, it has to be an equal sign not a less than/greater than sign so pyth. is already out

if a^2 + b^2 > c^2
c^2 = a^2 + b^2 - 2ac*cos(theta)

we can plug in the formula a^2 + b^2 - 2ac*cos(theta) into the first equation to get

a^2 + b^2 - 2ac*cos(theta) < a^2 + b^2

try cancelling out like terms on both sides of the inequality and solve the inequality for cos(theta)

Answer 48

a^2 + b^2 - 2ac*cos(theta) < a^2 + b^2

if you take away the a^2 and b^2 terms from both sides you get

-2ac*cos(theta) < 0

what happens when you divide both sides by -2ac? remember the rules for inequality operations

Answer 49

0-2ac

Answer 50

on the right side

Answer 51

-2ac is being muliplied not subtracted, so dividing by -2ac would leave us with

cos(theta)>0

since the inequality flips directions when you multiply/divide by a neg. number

anyway, if cos(theta) > 0 then is theta acute or obtuse?

Answer 52

um

Answer 53

acute

Answer 54

good

so cos(theta) > 0 and theta = acute (A+D)

Answer 55

as a hint try solving for 2abcos(theta) first, then plug in a, b, and c as needed

Answer 56

C

Answer 57

B?

Answer 58

a^2 +b^2 - 2abcos(theta) = c^2

therefore

2abcos(theta) = a^2 + b^2 - c^2 = ?

Answer 59

D

Answer 60

:/

Answer 61

well done, D