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Projeck
Registered User regular

I have the right answer, just wondering why it's correct

So, I'm supposed to solve this triangle using the law of sines and law of cosines

given A = 50° b = 15 and c = 30

so I plug that into the cos formula :

a^2 = 900 + 225 - 2 (450) cos50° -> some algebra -> a = ~23.38

with that, I go ahead and use the sin formula to find C

23.377 / sin 50° = 30 / sin C -> C = ~79.44°

Then I can find B by subtracting from 180°, and B = ~50.56°

but, that's wrong, (I can check in the back of the book, side a is correct, though) and if I find B and subtract from 180, I get the correct angles

why? (sorry, my teacher is terrible and doesn't explain things)

edit: the correct angles are B = ~29.44° and C = ~100.56°

So, I'm supposed to solve this triangle using the law of sines and law of cosines

given A = 50° b = 15 and c = 30

so I plug that into the cos formula :

a^2 = 900 + 225 - 2 (450) cos50° -> some algebra -> a = ~23.38

with that, I go ahead and use the sin formula to find C

23.377 / sin 50° = 30 / sin C -> C = ~79.44°

Then I can find B by subtracting from 180°, and B = ~50.56°

but, that's wrong, (I can check in the back of the book, side a is correct, though) and if I find B and subtract from 180, I get the correct angles

why? (sorry, my teacher is terrible and doesn't explain things)

edit: the correct angles are B = ~29.44° and C = ~100.56°

0

## Posts

travathianonProjeckonAwkonProjeckonNote that this function is not one to one, and that there are multiple angles

xwhich will give the samesin xvalue. So the inverse of the sine, the arcsine, is in fact not a function.For your particular problem, your calculator is only giving you one of the possible angles of C which would give you the correct sin C. Note that sin(79.44 degrees) = 0.983063532 = sin(100.56 degrees).

One way this should jump out at you is that if you center yourself at 90 degrees the sine function looks symmetric, and 79.44 and 100.56 degrees have the same absolute distance from 90 degrees. Both angles are potentially valid for being in a triangle, so you'll have to be careful and check yourself when using the inverses of trigonometric functions.

SavantonAwkonseems relevant

nevilleonedit: also, nope

Projeckon