Two corners of a triangle have angles of # (5 pi )/ 8 # and # ( pi ) / 2 #. If one side of the triangle has a length of # 6 #, what is the longest possible perimeter of the triangle?

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Answer 1 · 2017-12-02T09:06:31

Perimeter #= a + b + c = color (green)(36.1631)#

Sum of the three angles of a triangle is equal to #180^0 or pi#

As the sum of the given two angles is #=(9pi)/8# which is greater than #pi#, the given sum needs correction.

It’s assumed that the two angles be #color(red)((3pi)/8 & pi/2)#

#/_A = (5pi)/8, /_B = pi/2,#
# /_C = pi -( ((3pi)/8 )- (pi/2)) =pi - (7pi)/8 = pi/8#

To get the longest perimeter, length 6 should correspond to the smallest #/_C = pi/8#

#a / sin( /_A )= b / sin( /_B )= c / sin( /_C)#

#a / sin ((3pi)/8) = b / sin (pi/2) = 6 / sin (pi/8)#

#a = (6 * sin ((3pi)/8)) / sin (pi /8)#
#a = (6 * 0.9239)/ 0.3827 =color(blue)( 14.485)#

#b = (6 * sin (pi/2))/ sin (pi/8)#
#b = 6 / 0.3827 = color(blue)(15.6781)#

Perimeter #= a + b + c = 6 + 14.485 + 15.6781 = color (green)(36.1631)#