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

1 Answer
Somebody N.
Apr 14, 2018

See below.

Explanation:

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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

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