trig question

[Deviouspro]

i got stuck on this for a while
i got all the answers using law of sin/cos

Solve for all possible missing parts of triangle ABC, if possible. If not possible, turn in an explanation as to why.

a = 8
b = 6
B = 40 degrees

but... the missing side i got was 10.7... etc
i thought it would be 10 exact

[Zaczaco11]

i dont think thats accurately solvable in the way that you listed it because, if i remember correctly, you need to know if A and B are opposites of eachother and what kind of triangle it is...but its been almost 3 years since i've studied this so i could be wrong.

[TheBBQ]

i got stuck on this for a while
i got all the answers using law of sin/cos

Solve for all possible missing parts of triangle ABC, if possible. If not possible, turn in an explanation as to why.

a = 8
b = 6
B = 40 degrees

but... the missing side i got was 10.7... etc
i thought it would be 10 exact[/b]

Should be solvable. Let's assume capital letters are the values of the angle.
Use the sine law to find the value of A and from that you can find the value of C by doing subtracting. 180-40-A = C
From this you can use the sin law once again to find the value of c.
Could be an ambiguous case though.

[festo]

Draw a picture of the triangle, will be much simpler to solve (Type of triangle being used)

[moemoneyman]

Nevermind

[ZumieZ]

Draw a picture of the triangle, will be much simpler to solve (Type of triangle being used)[/b]

Draw a picture of the triangle, will be much simpler to solve (Type of triangle being used)[/b]

hmm ye easier if diagram drawn

[Deviouspro]

nvm anwser is 9.2 close

[bknight]

prob coulda googled a program to just plug in the numbers and get he sides/angles :P.

but if i had a calculater i woulda helped you out. my ones with the functions are in school .

[EG4L]

haha, my bro has a proper expensive calculator (colour nd shiz) and he programmed loads of formulas so in the exam i just typed in the legnths etc and got the answers

[Deviouspro]

there were 2 triangles possible
fuck -__-

[agendas]

SOH CAH TOA my friend

[tHeUnBeAtAbLe]

6/sin(40) = 8/sin(A) ---> sin(B) = (8*sin(40))/6 ---> b = 58.98 ~ 59 degrees

180 - sin(A) - sin(B) = 180 - 40 - 59 = 81 degrees = sin©

6/sin(40) = c/sin(81) ---> c = (6*sin(81))/sin(40) ---> 9.219 ~ 9.2

[Deviouspro]

SOH CAH TOA my friend [/b]

lol that doest work here
u have to use law os sines or law os cosines
which is what theunbeatable did and i did
but there was other triangle possible because the triangle was ambiguous
the angle could be obtuse
h34r:

[TheBBQ]

I'm a genius.
Just draw it out next time and it'll be ez pz.

[tHeUnBeAtAbLe]

I'm a genius.
Just draw it out next time and it'll be ez pz.[/b]

it's not exactly using Pythagoras theorem... so you really can't do shit from just drawing it