A triangle has sides A, B, and C. If the angle between sides A and B is #(pi)/8#, the angle between sides B and C is #(pi)/2#, and the length of B is 3, what is the area of the triangle?

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Answer 1 · 2018-06-14T11:16:03

Area of the triangle is #1.86# sq.unit.

Angle between sides # A and B# is # /_c= pi/8=180/8=22.5^0#

Angle between sides # B and C# is # /_a= pi/2=180/2=90^0 :.#

Angle between sides #C and A# is

# /_b= 180-(22.5+90)=67.5^0# The sine rule states if

#A, B and C# are the lengths of the sides and opposite angles

are #a, b and c# in a triangle, then, #A/sina = B/sinb=C/sinc#

# B=3 :. A/sin a=B/sin b :. A/sin 90=3/sin 67.5#

#:. A= 3 * (sin 90/sin 67.5) ~~ 3.25 (2dp) #

Now we know sides #A~~3.25 , B=3# and their included angle

#/_c = 22.5^0#. Area of the triangle is #A_t=(A*B*sin c)/2#

#:.A_t=(3.25*3*sin 22.5)/2 ~~ 1.86# sq.unit

Area of the triangle is #1.86# sq.unit [Ans]