A triangle has sides A, B, and C. If the angle between sides A and B is #(pi)/6#, the angle between sides B and C is #(7pi)/12#, and the length of B is 11, what is the area of the triangle?

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Answer 1 · 2016-02-13T18:46:50

Find all 3 sides through the use of law of sines, then use Heron's formula to find the Area.

#Area=41.322#

The sum of angles:

#hat(AB)+hat(BC)+hat(AC)=π#

#π/6-(7π)/12+hat(AC)=π#

#hat(AC)=π-π/6-(7π)/12#

#hat(AC)=(12π-2π-7π)/12#

#hat(AC)=(3π)/12#

#hat(AC)=π/4#

Law of sines

#A/sin(hat(BC))=B/sin(hat(AC))=C/sin(hat(AB))#

So you can find sides #A# and #C#

Side A

#A/sin(hat(BC))=B/sin(hat(AC))#

#A=B/sin(hat(AC))*sin(hat(BC))#

#A=11/sin(π/4)*sin((7π)/12)#

#A=15.026#

Side C

#B/sin(hat(AC))=C/sin(hat(AB))#

#C=B/sin(hat(AC))*sin(hat(AB))#

#C=11/sin(π/4)*sin(π/6)#

#C=11/(sqrt(2)/2)*1/2#

#C=11/sqrt(2)#

#C=7.778#

Area

From Heron's formula:

#s=(A+B+C)/2#

#s=(15.026+11+7,778)/2#

#s=16.902#

#Area=sqrt(s(s-A)(s-B)(s-C))#

#Area=sqrt(16.902*(16.902-15.026)(16.902-11)(16.902-7.778))#

#Area=41.322#