Question

For questions 11 - 14, solve the given triangles by finding the missing angle and other side lengths.

Answer 1

sorry if this takes a while, people on QC are being butts as usual

Answer 2

anyway, for 11 you can use law of sines to solve for side B

Answer 3

b/sin(B)=c/sin(C)
b/sin(75)=7.8/

Answer 4

ugh im confused

Answer 5

you don't have angle C or side c so we can't use those in the formula

but we can use a and b, A and B, to find b

b/sin(B) = a/sin(A)

then you can solve for b

Answer 6

b/sin(B) = a/sin(A)
b/sin(75)=7.8/sin(48)
b=10.14

Answer 7

cool

now, since we have angle A and B, we can simply use the 180 - angle A - angle B rule to find angle C

180 - 75 - 48 = 57 for angle C

then use law of sines again to find c

Answer 8

c^2=a^2+b^2-2*a*b cos(C)
c^2=7.8^2+10.14^2-2*7.8*10.14 (cos57)
c^2=77.51
c=8.80

Answer 9

that's law of cosines

Answer 10

oh shoot

Answer 11

lol hold on im sorry

Answer 12

I mean, it'll probably work either way, but just to double check

c/ (cos57) = 7.8/cos(48)

Answer 13

6.35

Answer 14

huh that's weird

anyway I'm going to check to see which one we should use, law of sines vs cosines

Answer 15

okay sounds good il be here

Answer 16

apparently in terms of accuracy, law of cosines takes precedence over law of sines

so I guess let's go with 8.80 for side c

Answer 17

anyway for 12) we only have one side so we have to start with law of sines

first let's start by finding angle B = 180 - angle A - angle C = 122

then let's try finding side (a)

b/sin(B) = a/sin(A)

Answer 18

200/sin(21)=a/sin(37)

Answer 19

you used the side 200 which is across from angle B
angle B is 122 not 21

so 200/sin(122) = a/sin(37) then find a

Answer 20

141.93 =a

Answer 21

awesome

then you can use law of cosines to find side c

Answer 22

c²=a²+b²-2abCos(C)
c^2=141.93^2+200^2-2*141.93*200 (cosC)

Answer 23

* 21 degrees

Answer 24

good but angle C = 21 degrees, plug that in and find c

Answer 25

c²=a²+b²-2abCos(C)
c^2=141.93^2+200^2-2*141.93*200 (cos21)
c^2=7142.9
c=84.52

Answer 26

good, that's the triangle solved

Answer 27

anway for #13 let's label angle C and angle A on the triangle:

Answer 28

first let's find angle B by subtracting 180 - angle C - angle A = 180 - 18 - 140 = 22 degrees

Answer 29

oh okay so now we have all the angles

Answer 30

A=140
B=22
C=18

Answer 31

(got momentarily kicked from the site)

anyway, since we only have 1 side, we have to use law of sines first to find side b

b/sin(B) = a/sin(A) as usual, find b

Answer 32

* take a look at the diagram:

angle A = 18
angle B = 22
angle C = 140

Answer 33

b/sin(B) = a/sin(A)
b/sin(22)=1/sin(128)

Answer 34

b=0.48

Answer 35

b/sin(22)=1/sin(18)

18 not 128

Answer 36

b=1.21

Answer 37

awesome now we got a, b, and all the angles, then use law of cosines to find c

Answer 38

c²=a²+b²-2abCos(C)
c^2=1^2+1.21^2-2*1*1.21 (cos140)
c^2=4.32
c=2.08

Answer 39

awesome so that's triangle 13solved

Answer 40

so im trying to do #14 on my own. Im using #12 as a reference. But how did you get 122 in the first line

Answer 41

angle B = 180 - 21 - 37 = 122

Answer 42

14/sin(28) = a/sin(37)
a=17.95

Answer 43

@Vocaloid i tagged you, so whenever you are back on we can continue

Answer 44

oh I didn't realize that there was a #14 whoops

anyway the left side is good

14/sin(28)

but side a is across from 112 so it would be

14/sin(28) = a/sin(112)

Answer 45

14/sin(28) = a/sin(37)
a=17.95

14/sin(28)

14/sin(28) = a/sin(112)
a=27.65

Answer 46

right

Answer 47

good, then you'd use law of cosines to find angle C

Answer 48

*side c

Answer 49

what is b again

Answer 50

the side across from angle B

Answer 51

14

Answer 52

?*

Answer 53

yes

Answer 54

c^2=a^2+b^2-2ab cosC
c^2=27.65^2+14^2-2*27.65+14 (cos40)
c^2=915.95
c=30.26

Answer 55

that plus sign between the 27.65 and the 14cos(40) shouldn't be there

Answer 56

c^2=a^2+b^2-2ab cosC
c^2=27.65^2+14^2-2*27.65*14 (cos40)
c^2=367.45
c=19.17

Answer 57

yup that's it

Answer 58

thankssss