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 #(5pi)/12#, and the length of B is 2, what is the area of the triangle?

1 Answer
Jan 3, 2016

#Area=1.93184# square units

Explanation:

First of all let me denote the sides with small letters a, b and c
Let me name the angle between side "a" and "b" by #/_ C#, angle between side "b" and "c" #/_ A# and angle between side "c" and "a" by #/_ B#.

Note:- the sign #/_# is read as "angle".
We are given with #/_C# and #/_A#. We can calculate #/_B# by using the fact that the sum of any triangles' interior angels is pi radian.
#implies /_A+/_B+/_C=pi#
#implies pi/6+/_B+(5pi)/12=pi#
#implies/_B=pi-(7pi)/12=(5pi)/12#
#implies /_B=(5pi)/12#

It is given that side #b=2.#

Using Law of Sines
#(Sin/_B)/b=(sin/_C)/c#
#implies (Sin((5pi)/12))/2=sin((5pi)/12)/c#
#implies 1/2=1/c#
#implies c=2#

Therefore, side #c=2#

Area is also given by
#Area=1/2bcSin/_A=1/2*2*2Sin((7pi)/12)=2*0.96592=1.93184#square units
#implies Area=1.93184# square units