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Anonymous Thu 29 Oct 2015 21:08:33 No.7624984 View Reply Report Delete

This may be the most basic bitch question ever posted here: how come when finding all sides and angles on a triangle using law of sines I end up getting different values with using the law to find the third angle, and just finishing out the problem with 180 - (A+B) = C?

Solve the triangle.
a = 5, b = 6, C = 30deg

I use law of cos to get c = 3.01
then I use law of sin for A, I get A = 56.2deg

180 - (56.2+30) = 93.8deg for the final angle right?

Then why when I do law of sin for B, I get B = 85.3?

pic related

Anonymous

Anonymous Thu 29 Oct 2015 21:14:34 No.7624997 Report Delete

Quoted By: >>7625025

>>7624984
It would help if you defined your question more carefully.
Sides a, b are what?
Is a or b the hypotenuse, is it adjacent or opposite to angle C?
Is there even a hypotenuse that is unique?
It is quite obvious that you mean that A is opposite a, B is opposite b and C is opposite c, but there are still many other triangles that can be created.
A picture or exact description would help.

Anonymous

Anonymous Thu 29 Oct 2015 21:21:39 No.7625025 Report Delete

Quoted By: >>7625104

>>7624997
>Solve the triangle.
>a = 5, b = 6, C = 30deg
This is the exact question given, no picture. This lesson is on finding missing components of triangles using law of cosines and law of sines, so it's assumed that there is no hypotenuse, it's not a right triangle, so no need for SOH CAH TOA or anything. A is opposite a, B is opposite b and C is opposite c, with lowercase being sides and upper being angles. The given info was an SAS triangle, meaning side and adjacent side, with the angle those two sides form. Does that clear it up? Sorry for bad formatting.

Anonymous

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Anonymous Thu 29 Oct 2015 21:51:50 No.7625097 Report Delete

Quoted By:

Anonymous

Anonymous Thu 29 Oct 2015 21:54:53 No.7625104 Report Delete

Quoted By: >>7625132 >>7625267

>>7625025
What we know for certain is:
a = 5, b = 6, c = 3.01.
We also know that A = 56.2 deg and that C = 30 deg.
We wish to apply the sine rule to find angle B.
Let's use the sine rule in as follows:
(sin B)/b = (sin C)/c.
Alternatively, we could use angle A and side a instead of angle C and side c.
Let's draw a picture.
So what we have a sides b, c and known angle C.
But angle C is not between b, c.
i.e. the triangle described by sides b, c and angle C is not unique.
Which is why you're getting two different answers.
We can also show that this is the case algebraically.
Going back to our equation, we have (rearranged):
sin B = (b(sin C))/c.
This gives sin B = 0.9978692864.
Now, draw the graph for sine.
Let's look at the first branch of the curve (so from 0 to 360 degrees).
Where does the graph take the value 0.9978692864?
At two places.
This means that when you apply the inverse, which only exists for bijective functions (and hence we take, by convention, the inverse of the primary branch, i.e. from -180 to 180 degrees), you will get the smaller of the two values.
This is why you're getting B = 85.3.
To obtain the second possible (and in this case, correct), we simply do 180 - 85.3 (which is obvious when you look at the graph).

Anonymous

Anonymous Thu 29 Oct 2015 22:04:43 No.7625132 Report Delete

Quoted By: >>7625267 >>7625296

>>7625104
Sorry, I forgot to define what a bijective function is for you.
A bijective function is a function that is both injective and surjective.
Let f: A --> B be a function.
Then f is injective if, for all x, y in A, f(x) = f(y) implies that x = y.
So, if we look at sine, we see that from 0 to 360 degrees, the function is NOT injective (there are two values of x at which sin(x) = 0.5 for example).
Additionally, f is surjective if, exists an x in A such that f(x) = y, for all y in B.
For sine, the range of our function is [-1, 1].
We can easily see why we take the branch that we do for sine.

Anonymous

Anonymous Thu 29 Oct 2015 22:58:19 No.7625267 Report Delete

Quoted By: >>7625282

>>7625104
>>7625132
Ahh that makes sense. I asked a couple of my friends in higher level classes and they didn't know either, so thanks a lot for the in-depth reply.

Anonymous

Anonymous Thu 29 Oct 2015 23:09:34 No.7625282 Report Delete

Quoted By:

>>7625267
No problem.
I hope this has made you realise that you can't just blindly apply rules (no matter how elementary they are).

Anonymous

Anonymous Thu 29 Oct 2015 23:25:25 No.7625296 Report Delete

Quoted By: >>7625301

>>7625132
is bijective the same as one-to-one?

Anonymous

Anonymous Thu 29 Oct 2015 23:27:59 No.7625301 Report Delete

Quoted By: >>7625369

>>7625296
Injective is sometimes called one-to-one.
Surjective is sometimes called onto.
Bijective is sometimes called a one-to-one correspondence.

Anonymous

Anonymous Fri 30 Oct 2015 00:08:04 No.7625369 Report Delete

Quoted By: >>7625377

>>7625301
A bijection can be called "one to one" whenever just like it can be called "onto"
People generally restrict "one to one" to injections though.

Anonymous

Anonymous Fri 30 Oct 2015 00:12:04 No.7625377 Report Delete

Quoted By:

>>7625369
Hence why I said sometimes, but yeah, different authors have different preferences.

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