A triangle has two corners with angles of # pi / 6 # and # (3 pi )/ 8 #. If one side of the triangle has a length of #8 #, what is the largest possible area of the triangle?

Question

1169 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-19T06:43:33

The area is #=2.75#

enter image source here

The angles ot the triangle are

#hatA=1/6pi#

#hatB=3/8pi#

#hatC=pi-(1/6pi+3/8pi)=pi-13/24pi=11/24pi#

The side of length is opposite the smallest angle in the triangle

So,

#a=6#

We apply the sine rule to the triangle

#b/sin hatB=a/sin hatA#

#b/sin(3/8pi)=6/sin(1/6pi)#

#b=6*sin(3/8pi)/sin(1/6pi)=11.1#

The area of the triangle is

#area =1/2absin hatC=1/2*6*11.1*sin(11/24pi)#

#=2.75#