A triangle has sides A, B, and C. The angle between sides A and B is #(pi)/3# and the angle between sides B and C is #pi/6#. If side B has a length of 11, what is the area of the triangle?

1 Answer
Alan P.
Nov 19, 2016

#"Area"_triangle= (121sqrt(3))/8# (using a calculator: #~~26.197#)

Explanation:

Since #/_ (A:B)= pi/3# and #/_(B:C)=pi/6#

#rarr /_(A:C) = pi-(pi/3+pi/6)=pi/2#
and the triangle is a right angled triangle:
enter image source here

#A/B = cos(pi/3) = 1/2#
#color(white)("XXX")rarr A= 1/2xxB = 11/2#

#C/B=sin(pi/3)= sqrt(3)/2#
#color(white)("XXX")rarr C= sqrt(3)/2xxB = (11sqrt(3)/2)#

#"Area"_triangle = 1/2* "base" * "height" = 1/2 xx 11/2 xx (11 sqrt(3))/2 = (121sqrt(3))/8#

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