Question

Use the Law of sines to find the missing angle of the triangle

Find m< B to the nearest tenth.
http://imgur.com/0Vxt8tc

Answer 1

@johnweldon1993 help with this one?

Answer 2

Alright so here...we know Angle A...and side a....and side c...and FIRST we want angle C

same equation, little different process..

\[\large \frac{\sin(A)}{a} = \frac{\sin(C)}{c}\]

plug in what we have

\[\large \frac{\sin(40)}{13} = \frac{\sin(C)}{19}\]

So can you solve that for sin(C)?

Answer 3

is it 70 degrees?

Answer 4

Rounded up to 70 yes...perfect..

Okay now...let me draw something quick...

Answer 5

What is m<B?

Answer 6

70?

Answer 7

Perfect :)

Answer 8

since you here can you help with another? haha I hate trig

Answer 9

haha of course :)

Answer 10

omg thanks so much overwhelmed with algebra lol

Answer 11

1. find m<B, given a=11, b=12, and c=17.

Answer 12

Alright...so here....3 sides...no angles are know...Seems like a law of COSINES problem...have you learned that yet?

Answer 13

lol i tried but failed miserably

Answer 14

isnt cos something like cos=a/h?

Answer 15

Lol :P alright...well we want angle B right?

Answer 16

Well...that is what COS is...but that only applies to right triangles...

Answer 17

yupppp

Answer 18

However...the law of cosines states

\[\large b^2 = a^2 + c^2 - 2ac\cos(B))\]

Answer 19

I wrote it like that to solve the the angle B here.....you can write it to solve for the other angles as well...

to solve for angle A we would have:

\[\large a^2 = b^2 + c^2 - 2bc\cos(A)\]

and to solve for angle C we would have

\[\large c^2 = a^2 + b^2 - 2ab\cos(C)\]

Answer 20

does that looks familiar at least? lol

Answer 21

okay so i should have 12^2=11^2+17^2-2abcos(B)?

Answer 22

sortaaa lol

Answer 23

i mean accos

Answer 24

well you CAN write it like that...however I like to do the algebra first to make it look better...we are solving for B so lets isolate the cos(B)

\[\large \cos(B) = -\frac{(b^2 - a^2 - c^2)}{2ac}\]

make sense there?

Answer 25

And then to solve for B we just do the arccos of the whole thing

\[\large \cos^{-1}(-\frac{b^2 - a^2 - c^2}{2ac})\]

Answer 26

okay

Answer 27

Didn't sound to reassuring lol...did that make sense?

Answer 28

lol yeah i guess

Answer 29

haha well no lets do it the way you did it originally :)

Answer 30

hahahaha lol sorry XD

Answer 31

No dont worry about it hun :)

\[\large 12^2=11^2+17^2-2(11)(17)cos(B)\]

Okay look good? Just plugged in all the numbers

Answer 32

looks tasty

Answer 33

Haha alright then ;P haha

So how would you go about solving this?

Answer 34

exponents first lol

Answer 35

mmhmm.....so then we have?

Answer 36

then multiply the 2 and so on lol

Answer 37

144=121+289-374cos B

Answer 38

Right...so lets begin that process...

\[\large 144 = 121 + 289 - 374cos(B)\]

Next....? :)

Answer 39

lol 144=410-374cosB

Answer 40

And of course OS had to lagg big time

regardless...yes that is correct...and next you would subtract the 410 from both sides...

\[\large -266 = -374\cos(B)\]

And then divide both sides by -374

\[\large cos(B) = .71122995\]

and take the arccos to find that

\[\large B = 44.66\]

and to the nearest tenth would be \[\large B = 44.7\]