How do you solve a triangle with two solutions? The problem says that A=40 degrees, b=10, and a=14. So I did this:

14>(10)(sin 40)
And this means that there are two solutions according to the equations my teacher gave me.

What do I do now?

by Sine Law :

sinB/b = sinA/a
sinB/10 = sin40/14
sinB = .459134
angle B = 27.33º or 180-27.33º

so B is either 27.33 or 152.67º

draw the two possible triangles, and use the sine law again to find c.

tysm

Oh, I see. So there would be two solutions for both B and C. I would just solve it like a regular triangle, and B' and C' would be 180 minus the values of B and C. Thanks!

You got it !!

If this is true... then how come if I have A=30, C=98, a=51... do i only end up with one solution?

To solve a triangle, you typically use the Law of Sines or the Law of Cosines. In this case, it seems like you attempted to use the Law of Sines. However, the inequality you wrote is incorrect.

The Law of Sines states that in any triangle, the ratio of the length of a side to the sine of its opposite angle is constant. This can be written as:

a / sin(A) = b / sin(B) = c / sin(C)

Given that A = 40 degrees, b = 10, and a = 14, you can rewrite the equation as:

14 / sin(40°) = 10 / sin(B)

To find angle B, you can solve for sin(B) by cross-multiplying:

14 * sin(B) = 10 * sin(40°)

Next, you need to isolate sin(B), so divide both sides of the equation by 14:

sin(B) = (10 * sin(40°)) / 14

Now, you can use the arcsin function (also known as sin^(-1)) to find the value of B:

B = arcsin((10 * sin(40°)) / 14)

Using a calculator, evaluate the right side of the equation and find the two possible values of B. These will be your two solutions.

Keep in mind that in this case, you are dealing with an ambiguous case where there are two possible triangles that satisfy the given information. To fully solve the triangle, you will need more information, such as another angle or side length.