A triangle has two corners with angles of # pi / 12 # and # pi / 8 #. If one side of the triangle has a length of #1 #, what is the largest possible area of the triangle?

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Answer 1 · 2017-12-02T13:15:50

Largest possible area of the triangle # = color(blue)(0.45)#

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

#/_A = (pi)/12, /_B = pi/8,#
# /_C = pi -(( pi/12) + (pi/8)) =pi - (5pi)/24 = (19pi)/24#

To get the largest possible area, length 1 should correspond to the smallest #/_A = pi/12#

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

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

#b = (1 * sin ((pi)/8)) / sin (pi /12)#
#b = 0.3827 / 0.2588 =color(blue)( 1.4787)#

#c = (1*sin (19pi/24))/ sin (pi/12)#
#b = 0.6088 / 0.2588 = color(blue)( 2.3524)#

Semi-Perimeter #s = (a + b + c )/2 =( 1 + 1.4787 + 2.3524 )/2= color (green)(2.4156)#

#s - a = 2.4156 - 1 = 1.4156#
#s - b = 2.4156 - 1.4787 = 0.9369#
#s - c = 2.4156 - 2.3524 = 0.0632#

Area of #Delta ABC = sqrt(s (s-a) (s - b) (s - c))#
# = sqrt ( 2.4156 * 1.4156 * 0.9369 * 0.0632) = 0.45#