A triangle has sides A, B, and C. The angle between sides A and B is #(5pi)/6# and the angle between sides B and C is #pi/12#. If side B has a length of 1, what is the area of the triangle?

1 Answer
Costas C.
Jan 3, 2016

Sum of angles gives an isosceles triangle. Half of the enter side is calculated from #cos# and the height from #sin#. Area is found like that of a square (two triangles).

#Area=1/4#

Explanation:

The sum of all triangles in degrees is #180^o# in degrees or #π# in radians. Therefore:

#a+b+c=π#

#π/12+x+(5π)/6=π#

#x=π-π/12-(5π)/6#

#x=(12π)/12-π/12-(10π)/12#

#x=π/12#

We notice that the angles #a=b#. This means that the triangle is isosceles, which leads to #B=A=1#. The following image shows how the height opposite of #c# can be calculated:

Image

For the #b# angle:

#sin15^o=h/A#

#h=A*sin15#

#h=sin15#

To calculate half of the #C#:

#cos15^o=(C/2)/A#

#(C/2)=A*cos15^o#

#(C/2)=cos15^o#

Therefore, the area can be calculated via the area of the square formed, as shown in the following image:

1
2
3
4
5

#Area=h*(C/2)#

#Area=sin15*cos15#

Since we know that:

#sin(2a)=2sinacosa#

#sinacosa=sin(2a)/2#

So, finally:

#Area=sin15*cos15#

#Area=sin(2*15)/2#

#Area=sin30/2#

#Area=(1/2)/2#

#Area=1/4#

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