Question

3?

Answer 1

@Vocaloid

Answer 2

have you tried using law of cosines? by finding the missing sides/angles it should be clear whether the triangle(s) are possible or not

Answer 3

0?

Answer 4

good, 0, since angle A is the biggest angle then a must be the biggest side which is not the case here

Answer 5

c²=a²+b²-2abCos(C)

Answer 6

good, so what do you get when you start plugging values into the law of cosines?

Answer 7

A=10.2
B=9.8
C=5.7

Answer 8

just want to ge tthe values straight

Answer 9

good, now solve for cos(C)

Answer 10

c²=a²+b²-2abCos(C)
5.7^2=10.2^2+9.8^2-2*10.2*9.8(CosC)

Answer 11

Choice C

Answer 12

well done

Answer 13

any attempts to calculate the projection?

Answer 14

notice how the dot product is in the numerator, start with the dot product of your two vectors

Answer 15

remember the rule for dot product of two vectors:

1. multiply the x-coordinates
2. multiply the y-coordinates
3. add the results from step 1 and 2

Answer 16

10 + -6 = 4

Answer 17

good, now calculate the magnitude of the vector being projected onto, in other words the magnitude of (5,-1)

Answer 18

the magnitude is multiplication right

Answer 19

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

Answer 20

sqrt 26

Answer 21

good, now you have the dot product and magntiude, plug these into the projection formula to find the projection

Answer 22

\[\frac{ \sqrt{26} * 4 }{ 4 }\]

Answer 23

\[\frac{ 4 * (5,-1) }{ \sqrt{26}^2 }\]

Answer 24

good, now what do you get when you simplify the 4/sqrt(26)^2 part

Answer 25

2/13

Answer 26

choice b

Answer 27

awesome

Answer 28

b^2=ac-2*a*c*(cos B)
b^2=22^2*14^2-2*22*14*(cos 110)

Answer 29

oh wait i use law of sines right

Answer 30

no it's law of cosines

Answer 31

let me just double check your calculations

Answer 32

okay

Answer 33

b^2=22^2+14^2-2*22*14*(cos 110)

there needed to be a + sign between the 22^2 and the 14^2 not a * sign

Answer 34

B

Answer 35

good

Answer 36

B

Answer 37

yeah that's good

Answer 38

t

Answer 39

good

Answer 40

law of sines right

Answer 41

I think so yeah

Answer 42

\[\frac{ \sin(B) }{ b }=\frac{ \sin(C) }{ c }\]

Answer 43

good, now plug in the values of angle B, angle C, and side c to solve for b

Answer 44

\[\frac{ Sin(85) }{ b }=\frac{ Sin(31) }{ 9.3 }\]

Answer 45

Chioce A

Answer 46

well done

Answer 47

B

Answer 48

good

Answer 49

D

Answer 50

hm. I guess you could start with the large triangle and use law of cosines to try and find the bottom side

Answer 51

Im stuck between a and d

Answer 52

try to use the 60 degree angle as the reference angle for the law of cosines, find the bottom side, then use law of cosines again to find angle B

keep as many decimal places as possible since the answers are all very close to each other

Answer 53

12.3^2 = 13^2 + x^2 - 2(x)(13)cos(60) solve for x, you will get two values, let's use the larger x-value for now since we are solving for the big triangle

Answer 54

11.45

Answer 55

good, let's keep a few more digits so let's use 11.4538

then we can plug that back into the law of cosines to find angle B

Answer 56

b^2=a^2+c^2-2*a*c*(Cos B)

Answer 57

good, start plugging in the sides and angle

Answer 58

b=11.4538
a=13
c=12.3

right?

Answer 59

the lower case letter matches the angle it is across from

so side b is the side across from angle B, so b = 13

technically we didn't assign an a and c value but it doesn't really matter here, we can just pick a = 12.3 and c = 11.4538

Answer 60

b^2=a^2+c^2-2*a*c*(Cos B)

13^2=12.3^2+11.4538^2-2*12.3*11.4538 (Cos B)

=0.4027

Answer 61

awesome, then you take the arccos of that to get angle B

Answer 62

1.16

Answer 63

your calculator is in radians

that would be 66.25 degrees which points us towards choice C

Answer 64

False

Answer 65

good

Answer 66

A, C, D, E,

Answer 67

AAA doesn't work; for law of cosines/sines you need at least one side
SAA: yes (law of sines)
SAS: yes (that's law of cosines)
SSS: yes (law of cosines)
SSA: maybe (my research says no?)
ASA: yes

so BCDF is my best attempt

Answer 68

B C

Answer 69

keep in mind pythaogrean theorem states that a^2 + b^2 = c^2 for a right triangle, it cannot be > or < than so A must be true and C false

B is correct

any thoughts on D? try to see which values of theta make cos(theta) < 0, or try drawing a triangle where c^2 > a^2 + b^2

Answer 70

yeah d is an answer

Answer 71

ii think

Answer 72

good so A+B+D

Answer 73

D

Answer 74

good