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?

1 Answer
Costas C.
Feb 13, 2016

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

Area=41.322

Explanation:

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

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