A triangle has two corners with angles of # (3 pi ) / 4 # and # ( pi )/ 6 #. If one side of the triangle has a length of #16 #, what is the largest possible area of the triangle?

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Answer 1 · 2018-02-15T07:22:31

Largest possible area of the triangle is #A_t ~~ color (blue)(174.85# sq units

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Given : #hatA = 3pi/4, B = pi/6,#

Third angle #hatC= pi - 3pi/4 -pi/6 = pi/12#

To get the longest area, length 16 should correspond to least angle #pi/12#

#a / sin A = b / sin B = c / sin C#

#a / sin((3pi)/4) = b / sin (pi/6) = 16 / sin (pi/12)#

# b = (16 * sin (pi/6)) / sin (pi/12) ~~ 30.91#

Area of triangle #A_t = (1/2) b c sin hatA = (1/2) * 30.91 * 16 sin ((3pi)/4) #

#=> 174.85# sq units