How do you solve the triangle given m∠B = 45°, a = 28, b = 27?

Start by drawing a diagram.

We know an angle and the side opposite this angle. We also know an additional side. Therefore, we will use the Law of Sines to solve this triangle, and we must be aware that this is an ambiguous case.

#sinA/a = sinB/b#

#sinA/28 = (sin45˚)/27#

#A = sin^-1((28sin(45˚))/27)#

#A = 47.2˚#

Let's check for the possibility of two triangles. If there was another triangle, the alternate #/_A# would measure #180 - 47.2 = 132.8˚#. The only question that remains is: Would an angle this size fit without having the sum of the angles inside the triangle surpass #180˚#.

The answer: yes. This is because #132.8˚ + 45˚ = 177.8˚#, which is inferior to #180˚#.

Now, let's find the two cases for #/_C#.

Case 1, where #/_A = 47.2˚#

#/_C = 180˚ - 47.2˚ - 45˚ = 87.8˚#

Case 2, where #/_A = 132.8˚#

#/_C = 180˚ - 132.8˚ -45˚ = 2.2˚#

The last step in this problem is to determine the two possible measures of side C.

Case 1, where #/_C = 87.8˚#

#sinB/b = sinC/c#

#(sin45˚)/27 = (sin87.8˚)/c#

#c ~= 38.2#

Case 2, where #/_C = 2.2˚#

#sinB/b = sinC/c#

#(sin45˚)/27 = (sin2.2˚)/c#

#c ~=1.5#

In summary:

There are two triangles possible.

Triangle 1:

#/_A ~= 47.2˚#

#/_C ~= 87.8˚#

#c ~= 38.2#

Triangle 2:

#/_A ~= 132.8˚#

#/_C ~= 2.2˚#

#c ~= 1.5#

Since this is my 1000th answer, I have included practice exercises en masse and a special image.

Practice exercises:

#1.# Solve the following triangles. Check for one solution, no solutions or two solutions.

a) #/_A = 108˚, b = 52, a = 41#

b) #/_B = 50˚, b = 22, c = 54#

c) #/_C = 28˚, b = 14, c = 16#

d) #/_C = 94˚, a = 26, c = 3#

Hopefully this helps, and good luck!