A triangle has two corners with angles of # pi / 12 # and # pi / 6 #. If one side of the triangle has a length of #8 #, what is the largest possible area of the triangle?

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Answer 1 · 2017-12-05T11:06:07

#Area of# largest possible #Delta = color(purple)(43.7113)#

Three angles are #pi/12, pi/6, (pi -( (pi/12)+ ((pi)/6)) =3pi/4#

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

To get the largest possible are, smallest angle should correspond to the side of length 7

#8 / sin (pi/12) = b / sin ((pi)/6) = c / sin ((3pi)/4)#

#b = (8*sin (pi/6)) / sin (pi / 12) = (8*(1/2)) / (0.2588) = 15.456#

#c = (8* sin ((3pi)/4)) / sin (pi/12) = (8 * (1/sqrt2))/(0.2588 )= 21.858#

Semi perimeter #s = (a + b + c) / 2 = (8 + 15.456 + 21.858)/2 = 22.657

#s-a = 22.657 - 8 = 14.657#
#s-b = 22.657 -15.456 = 7.201#
#s-c = 22.657 - 21.858 = 0..799#

#Area of Delta = sqrt (s (s-a) (s-b) (s-c))#

#Area of Delta = sqrt( 22.657 * 14.657 * 7.201 * 0.799)#
#Area of# largest possible #Delta = color (purple)(43.7113)#