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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