Let's work with the law of sines and cosines.

For example, let triangle ABC be an oblique triangle with little a = 10,

little b = 14 and little c = 6.

Let's solve the triangle.

Now, let's mark in the figure the information that's given.

We're given that little a = 10,

little b = 14 and little c = 6.

Now, this is what we refer to as the SSS case.

Which stands for side-side-side,

because we're given all three sides of the triangle, but no angles.

And therefore, since we are not given a matching angle-side pair,

we cannot start off by using the law of sines,

but we can however start off by using the law cosines.

Now, when we're in this SSS case and we're going to be using the law of cosines,

what we do first is we find

the largest angle to see if our triangle has any obtuse angles.

So, we'll find the largest angle first,

and we know what angle that is because it's across from the longest side.

And since little b here is the longest side,

that means capital B is the largest angle.

So, we'll be using this middle formula down here.

That is we have b squared which is 14 squared is equal to a squared,

which is 10 squared,

plus C squared, just 6 squared - 2,

times a, times c,

times the cosine of capital B.

Now, solving this equation for cosine of B,

it says that cosine of B = 10

squared + 6 squared - 14 squared,

all divided by 2,

times 10, times 6.

In doing this calculation,

we get that this is equal to negative 1/2,

which means that B is equal to inverse cosine of negative 1/2,

which is equal to 120 degrees.

So, let's mark that on our figure here.

B is 120 degrees,

and now we have a matching angle-side pair,

capital B and little b.

So, let's use the law of sines to find capital A.

That is we have sine of capital A divided by little a,

is equal to sine of capital B divided by little b,

or sine of capital A is equal to little a,

times sine of capital B divided by little b,

which is equal to 10 times sine of 120 degrees divided by 14,

and the sine of 120 degrees is equal to the square root of 3 over 2.

So, therefore this is equal to 10,

times the square root of 3,

divided by 2, still divided by fourteen,

which is equal to 5 square root of 3 divided by 14, and therefore,

A is equal to the inverse sine of 5 times the square root of 3 divided by 14,

and using our calculator,

we get that this is approximately 38.2132 degrees.

Now, there is another angle A that is

less than 180 degrees whose sine is equal to this five square root of three over 14,

and therefore can be a candidate for an angle in a triangle,

namely 180 degrees, minus this 38.2132.

However, that angle would be obtuse,

and since B is already obtuse,

there's no way that obtuse possibility for A would work because remember,

there can be at most one obtuse angle in a triangle.

Therefore A is this 32.2132 degrees.

So, let's write that in our figure up here,

A is approximately 38.2132 degrees.

Now, it still remains to find C,

but we can do so by using the fact that the angle measures

in a triangle add up to 180 degrees.

That is C is equal to 180 degrees minus A plus B,

which is approximately 180 degrees minus

38.2132 degrees plus 120 degrees,

which is equal to 21.7868 degrees.

So, let's mark that on our figure as well.

C is approximately 21.7868 degrees.

Now, looking over here on the left,

you might be wondering why we started off by finding the largest angle first.

So, let's say that instead of solving for the largest angle B first,

we decided to solve for A first.

So, we'd be using this first formula down here,

and if we solve this for cosine of A,

we'd get cosine of A is equal to b squared, plus c squared,

minus a squared divided by two times b, times c,

which is equal to 14 squared plus 6

squared minus 10 squared divided by 2 times 14,

times 6, which turns out to equal 11 / 14,