Question

Help

Answer 1

well you have two angles A and C, how would you find the missing angle B?

Answer 2

law of sines

Answer 3

remember, when you have two angles in a triangle,

180 - angle - other angle =

will give you the missing angle

Answer 4

oh okay so

180-42-83=55

Answer 5

yup

then for part II) you can use a/sinA = b/sinB to find a

Answer 6

a/sinA = b/sinB

175/42=b/83

Answer 7

2075/6

Answer 8

check again, angle B is not 83 degrees

Answer 9

also you need to include sin in the denominators

Answer 10

a/sinA = b/sinB

175/sin42=b/sin55

b=214.24

Answer 11

175 is side b not a

we are solving for side a

Answer 12

a/sin(42) = 175/sin(55)

Answer 13

142.95

Answer 14

A+

Answer 15

oh right I forgot it's asking for side c too

you can use law of cosines i guess

Answer 16

c^2 = a^2 + b^2 - 2ab * cos(C)

you have everything except c, solve for c

Answer 17

should be -2ab not -2ac

Answer 18

c²=a²+b²-2abCos(C)

c^2=142.95^2+175^2-2*142.95*175 (Cos(83)

c^2=44962.3
c=212.0

Answer 19

A+ well done

Answer 20

well same logic as the previous problem, we are given two angles, how would we find the missing angle A?

Answer 21

part I --

180-35-45=20

Answer 22

for part 2 are we solving for angle B?

Answer 23

I mean b

Answer 24

check your math again

180 - 45 - 35 = 100 not 20

for part II) yes, use law of sines to find side b

Answer 25

17/sin(100) = b/sin(45)

Answer 26

b=12.21

Answer 27

have to be very careful with your a, b, and c labels

Answer 28

a is always across angle A
b is always across angle B
c is always across angle C

Answer 29

so looking at your diagram

a/sin(A) = c/sin(C) = ?

Answer 30

a/sin(A) = c/sin(C)
a/sin(100)=17/sin(35)
a=29.2

Answer 31

good

then for part III) use law of cosines to find b

Answer 32

b²=a²+c²-2acCos(B)

b^2=29.2^2+17^2-2*29.2*17 (cos35)
b^2-328.39
b=18.12

Answer 33

A+ well done

Answer 34

part I

180-130-30=20

Answer 35

good, then use law of sines for part II to find d

Answer 36

d/sin130=35/sin30

Answer 37

ugh this :/

Answer 38

almost

d/sin(130) = 35/sin(20)

not 30degrees

Answer 39

d=78.39

Answer 40

good

then for part III you can find the missing side (from suzie to the boat) using law of cosines

Answer 41

there is only two parts to this question :/

Answer 42

oh ok cool

Answer 43

don't forget your units (Feet)

Answer 44

b²=a²+c²-2acCos(B)


21^2=22^2+16^2-2*22*16 (cos B)

Answer 45

The calc for some reason is not giving me an answer is 299 roght?

Answer 46

21^2=22^2+16^2-2*22*16*x

try solving for x, then take the arccos of the result

Answer 47

64.87

Answer 48

good

then for part II, law of sines to find C

Answer 49

b/sin(B)=c/sin(C)
21/sin(64.87)=16/Sin(C)
C=84.22 degrees

Answer 50

hm, try that again

21/sin(64.87)=16/x

try solving for x then take the arcsin of the result

Answer 51

42.84

Answer 52

good, then for part III

we have angle B and angle C so angle A = 180 - B - C

Answer 53

180 - 42.84 - 64.87 = ?

Answer 54

72.29

sorry I was afk

Answer 55

good

Answer 56

b²=a²+c²-2acCos(B)

18^2=

Answer 57

ugh im confused

Answer 58

we want to find side a we have angle A, side b, and side c

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

Answer 59

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

a^2=18^2+20^2-2*18*20 (Cos 95)

Answer 60

check your notation again

a^2=18^2+20^2-2*18*20*cos(95)

Answer 61

click the approximate form button

Answer 62

28.049

Answer 63

good so 28.049 feet

for part II: plug in d = r*t

where d is the distance, r is the speed of car A, t is time, solve for time

Answer 64

28.049=30*t

t=0.93

Answer 65

oh whoops that's car B so that would be part III

for part II use the distance of car A (18 miles) and the speed from car A (30 mph)

Answer 66

18=30*t

Answer 67

like that

Answer 68

good

Answer 69

t=3/5

Answer 70

for part II

Answer 71

good

for part III use the distance 28.049 feet and speed 45 mph

Answer 72

28.049=45*t
t=0.62

Answer 73

would it be seconds?

Answer 74

travelling 28 miles in 0.62 would be quite a feat

the speed is given in mph so hours not seconds

Answer 75

so 0.62 hours

Answer 76

good

so for part IV car B is greater by (0.62-0.60) hours

Answer 77

0.02

Answer 78

good