Question

How do you find all possible measures of B if `A = 30^\circ`, `a = 13`, `b = 15` for triangle ABC?

Answer 1

See below.

Explanation:

From the diagram you can see that we have an ambiguous case. This gives us to unique triangles:

ABC and ACD

Using the sine rule:

`sinA/a=sinB/b=sinC/c`

Solving for `/_ABC`

`sin(30)/13=sin(B)/15`

Rearranging:

`sin(B)=(15sin(30))/13=((15)(1/2))/13=15/26`

`B=arcsin(15/26)=35.23^@` 2 d.p.

But we also have another angle `/_ADC`

These are angles on a straight line so:

`/_ADC=180^@-arcsin(15/26)=144.77^@` 2 d.p.

Triangle ABC`[1]`

Using `B=35.23^@`

`/_ACB=180^@-(30^@+35.23^@)=114.77^@`

Side `c`:

`sin(30^@)/13=sin(114.77)/c`

`c=(13sin(114.77))/(sin(30^@))=23.61`

So triangle ABC`[1]`

`bbA=30^@`

`bb(B)=35.23^@`

`bbC=114.77^@`

`bba=13`

`bb(b)=15`

`bbc=23.61`

Triangle ABC`[2]`

Using `B=144.77^@`

`/_ACB`

`180^@-(114.77^@+30^@)=5.23^@`

side `c`

`sin(30^@)/13=sin(5.23^@)/c`

`c=(13sin(5.23^@))/(sin(30^@))=2.37`

So triangle ABC`[2]`

`bbA=30^@`

`bb(B)=144.77^@`

`bbC=5.23^@`

`bba=13`

`bb(b)=15`

`bbc=2.37`