solve triangle. b=14 c=18 angleB=50 degrees

Question

there are two cases to the problem. thats all i know.

Answer 1

Maria,

as you say, there is not one unique solution for this triangle.  Using Law of Sines, we get:

sin C = 18 * sin(50) / 14 = 0.98491

There are two angles which satisfy this constraint: C = 80.035 and C = 180 - 80.035 = 99.965

From these two answers we can derive the rest of the triangle by summing the angles to 180 degrees and then using Law of Sines to solve for the remaining side:

A = 49.965
B = 50
C = 80.035
a = 13.993
b = 14
c = 18

or

A = 30.035
B = 50
C = 99.965
a = 9.148
b = 14
c = 18

Answer 2

Hey Maria,

The reason there are actually two different possible answers is that you are given 2 sides of the triangle, but the angle you're given isn't between those sides.  This is exactly why you can't prove triangle congruence by using Angle, Side, Side (also, the acronym is naughty, haha).

To get both possible solutions, use the Law of Cosines (c² = a² + b² - 2ab*cos(C)).  Since you know 2 sides and an angle, set it up so that the only variable is the missing side; to make the letters match, use the version of the Law of Cosines that looks like "b² = a² + c² - 2ac*cos(B)".  This will give you a quadratic: 196 = 324 + x² -2*18*x*cos(50).  Make one side equal zero by subtracting the 196 over to the other side and use the quadratic function (x = (-b ± √(b^2 - 4ac))/(2a)) to get both answers.  I'll leave that part to you.

Best of luck!

Answer 3

Hi Maria, 

One way to solve this problem is to use the Law of Sines: sin(A)/a = sin(B)/b = sin(C)/c.  Here a, b, c are the lengths of the sides, and A, B, C are the angles across from (not touching) the sides a, b, c respectively.  

Since you are given b, c, and B, you could plug them into sin(B)/b = sin(C)/c to find C.  Having found C, use the fact that the angles of a triangle sum to 180 degrees to find A.  Then plug A, B, and b into sin(A)/a = sin(B)/b to find a.

Hope this helps!