A triangle has two corners with angles of # (3 pi ) / 4 # and # ( pi )/ 8 #. If one side of the triangle has a length of #12 #, what is the largest possible area of the triangle?

Question

1194 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-23T13:00:48

Largest possible area of the triangle is #50.91# sq.unit.

Angle between Sides # A and B# is

# /_c= (3pi)/4=(3*180)/4=135^0#

Angle between Sides # B and C# is # /_a= pi/8=180/8=22.5^0 :.#

Angle between Sides # C and A# is

# /_b= 180-(135+22.5)=22.5^0#

For largest area of triangle #12# should be smallest size i.e

opposite to the smallest angle #:. B=12#

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=12 :. A/sina=B/sinb# or

#A/sin22.5=12/sin22.5 :. A = 12* sin22.5/sin22.5 = 12 # unit

Now we know sides #A=12 , B=12# and their included angle

#/_c = 135^0#. Area of the triangle is #A_t=(A*B*sinc)/2#

#:.A_t=(12*12*sin135)/2 ~~ 50.91# sq.unit.

Largest possible area of the triangle is #50.91# sq.unit [Ans]