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

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Answer 1 · 2017-10-19T22:02:57

Largest possible area of triangle = 262.5667

Three angles are #pi/12, pi/8, (17pi)/24#

To get larges area, length 6 must correspond to least angle #pi/12#

#6/ sin (pi/12) = b / sin (pi/8) = c / sin ((19pi)/24)#

#b = (6 * sin (pi)/8) / sin (pi/12) = 8.8715#

#c = (6*sin ((19pi)/24)) / sin (pi/ 12) = 14.1124#

Semiperimeter #s = ( a + b + c)/2 = (6+8.8715+14.1124)/2 = 14.492#

#s-a = 14.492 - 6 = 8.492#
#s-b = 14.492 - 8.8715 = 5.6205#
#s-c = 14.492 - 14.1124 = 0.3796#

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

Area of #Delta = sqrt(14.492 * 8.492 * 5.6205 * 0.3796) = **262.5667**#