A triangle has sides A, B, and C. The angle between sides A and B is #(7pi)/12# and the angle between sides B and C is #pi/12#. If side B has a length of 12, what is the area of the triangle?

1 Answer
May 19, 2017

#A=24.053# #units^2#

Explanation:

#Acolor(white)(00)color(black)(pi/12)color(white)(0000000)acolor(white)(00)color(white)(0)#

#Bcolor(white)(00)color(white)(pi/1 3)color(white)(0000000)bcolor(white)(00)color(black)(12)#

#Ccolor(white)(00)color(black)((7pi)/12)color(white)(0000000)c color(white)(00)color(white)(0)#

Let's find the remaining angle, #B#:

#pi-pi/12-(7pi)/12# leaves us with #pi/3#

#Acolor(white)(00)color(black)(pi/12)color(white)(0000000)acolor(white)(00)color(white)(0)#

#Bcolor(white)(00)color(black)(pi/3)color(white)(.)color(white)(0000000)bcolor(white)(00)color(black)(12)#

#Ccolor(white)(00)color(black)((7pi)/12)color(white)(0000000)c color(white)(00)color(white)(0)#

Now we should use law of sines

#(sin(pi/3))/12=(sin(pi/12))/a#

#a~~3.59#

#(sin(pi/3))/12=(sin((7pi)/12))/c#

#c~~13.4#

Now, to find the area, we use the equation #A=(hxxb)/2#, where #h# is the height.

#color(white)(a)color(white)(- - - - - - - - - -)color(black)(/)color(black)(|)#
#color(white)(a)color(white)(- - - - - - - - -)color(black)(/)color(white)(-)color(black)(|)#
#color(white)(a)color(white)(- - - - - - - -)color(black)(/)color(white)(- - .)color(black)(|)#
#color(white)(a)color(white)(- - - - - - -)color(black)(/)color(white)(- - -0)color(black)(|)#
#color(white)(- - - -)color(black)(a)color(white)(00000)color(black)(/)color(white)(- - - -0)color(black)(|)color(black)(h)#
#color(white)(a)color(white)(- - - - -)color(black)(/)color(white)(- - - - -0)color(black)(|)#
#color(white)(a)color(white)(- - - -)color(black)(/)color(white)(- - - - - -0)color(black)(|)#
#color(white)(a)color(white)(- - -)color(black)(/)color(white)(- - - - - - -0)color(black)(|)#
#color(white)(a)color(white)(- -)color(black)(/)color(white)(- - - - - - - -0)color(black)(|)#
#color(white)(-)color(black)(/)color(black)(..)color(black)(B)color(black)(........................................)#

#a=3.59#, and the angle #B# is #pi/3#. We need to find the remaining length, #h#.

#sin(pi/3)=h/(3.59)#

#sin(pi/3)xx3.59=h#

#h=3.11#

Now we can find the area:

#A=(hxxb)/2#

#A=(3.59xx13.4)/2#

#A=24.053# #units^2#