A triangle has two corners with angles of # pi / 4 # and # (3 pi )/ 8 #. If one side of the triangle has a length of #7 #, what is the largest possible area of the triangle?

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Answer 1 · 2017-12-02T13:37:24

Largest possible area of the triangle # = color(blue)(29.5747)#

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

#/_A = (pi)/4 /_B = 3pi/8,#
# /_C = pi -(( pi/4) + (3pi/8)) =pi - (5pi/8) = (3pi)/8#

To get the largest possible area, length 1 should correspond to the smallest #/_A = pi/4#

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

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

#b = (7 * sin (3pi)/8) / sin (pi /4)#
#b = 6.4672 / 0.7071 =color(blue)(9.1461) #

#c = (7*sin (3pi)/8) / sin (pi/4)#
#c = color(blue)( 9.1461)#

Semi-Perimeter #s = (a + b + c )/2 =( 7 + 9.1461 + 9.1461)/2 = color (green)(12.6461)#

#s - a = 12.6461 - 7 = 5.6461#
#s - b = 12.6461 - 9.1461 = 3.5#
#s - c = 12.6461 - 9.1461 = 3.5#

Area of #Delta ABC = sqrt(s (s-a) (s - b) (s - c))#
# = sqrt ( 12.6461 * 5.6461 * 3.5 * 3.5) = 29.5747#