Two corners of a triangle have angles of # (5 pi )/ 8 # and # ( pi ) / 4 #. If one side of the triangle has a length of # 14 #, what is the longest possible perimeter of the triangle?

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Answer 1 · 2017-12-05T11:33:38

#Area of # largest possible #Delta = color (purple)(160.3294)#

Three angles are #pi/4, ((5pi)/8), (pi -( (pi/4)+ ((5pi)/8) =( pi/8)#

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

To get the largest possible are, smallest angle should correspond to the side of length 14

#14 / sin (pi/8) = b / sin ((pi)/4) = c / sin ((5pi)/8)#

#b = (14*sin (pi/4)) / sin (pi / 8) = (14*(1/sqrt2)) / (0.3827) = 25.8675#

#c = (14* sin ((5pi)/8) / sin ((pi)/8) = (14 * 0.9239)/(0.3827) = 33.7983#

Semi perimeter #s = (a + b + c) / 2 = (14+ 25.8675 + 33.7983)/2 = 36.8329#

#s-a = 36.8329 -14 = 22.8329#
#s-b = 36.8329 -25.8675 = 10.9654#
#s-c = 36.8329 - 33.7983 = 3.0346#

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

#Area of Delta = sqrt( 36.8329 * 22.8329 * 10.9654 * 3.0346)#
#Area of# largest possible #Delta = color (purple)(160.3294)#