A triangle has sides A,B, and C. If the angle between sides A and B is #(5pi)/12#, the angle between sides B and C is #pi/6#, and the length of B is 15, what is the area of the triangle?

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Answer 1 · 2017-06-27T06:16:35

The area of the triangle is #=56.25u^2#

The angle between #A# and #B# is

#=pi-(5/12pi+2/12pi)#

#=pi-(7/12pi)#

#=5/12pi#

So, the triangle is isoceles

The side #B# is #=15#

Therefore, the side #C# is #=15#

#sin(1/6pi)=1/2#

The area of the triangle is

#A=1/2*15*15sin(1/6pi)#

#=225/4=56.25#

Answer 2 · 2017-06-27T06:30:40

#area = 56.25# units squared

Here's all the info we know (I found that last angle by subtracting all the other angles from #2pi)#:
enter image source here

So, here's what we need to know for the area:
enter image source here

Once we find these values, we'll be able to use the formula #area=1/2(b xx h)#

Let's work on finding the height, #h#.

To do that, we just need to use #sin(pi/6)=h/15#, or #7.5=h#

Now we know the height, all that's left is to find the base, or #c#

First, let's put all our info in a table:

length #color(white)(0000)# angle
#A = ? color(white)(0000) A= pi/6#
#B = 15 color(white)(0000) B = (17pi)/12#
#C = ? color(white)(0000) A= (5pi)/12#

Let's use law of sines

#(sin((5pi)/12))/(C) = (sin((17pi)/12))/(15)#

#c = -15#, but because this problem is dealing in distances, we cannot have a negative length, so #c=15#

Now let's use our formula: #area=1/2(b xx h)#

#area = (15 xx 7.5)/2#

#area = 56.25# units squared