A triangle has sides A, B, and C. The angle between sides A and B is #pi/6# and the angle between sides B and C is #pi/12#. If side B has a length of 3, what is the area of the triangle?

1 Answer
Jan 8, 2016

#Area=0.8235# 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# by #/_ 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/12+/_B+(pi)/6=pi#
#implies/_B=pi-(pi/6+pi/12)=pi-(3pi)/12=pi-pi/4=(3pi)/4#
#implies /_B=(3pi)/4#

It is given that side #b=3.#

Using Law of Sines
#(Sin/_B)/b=(sin/_C)/c#
#implies (Sin((3pi)/4))/3=sin((pi)/6)/c#

#implies (1/sqrt2)/3=(1/2)/c#

#implies sqrt2/6=1/(2c)#

#implies c=6/(2sqrt2)#

#implies c=3/sqrt2#

Therefore, side #c=3/sqrt2#

Area is also given by
#Area=1/2bcSin/_A#

#implies Area=1/2*3*3/sqrt2Sin((pi)/12)=9/(2sqrt2)*0.2588=0.8235# square units
#implies Area=0.8235# square units