A triangle has two corners with angles of $( pi ) / 4$ and $( 7 pi )/ 12$ . If one side of the triangle has a length of $12$ , what is the largest possible area of the triangle?

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Answer 1 · 2018-03-02T12:19:32

Area of the largest possible triangle is $98.34$ sq.unit.

Angle between Sides $A and B$ is $/_c= pi/4=45^0$

Angle between Sides $B and C$ is $/_a= (7pi)/12=105^0 :.$

Angle between Sides $C and A$ is

$/_b= 180-(45+105)=30^0$

For largest area of triangle $12$ should be smallest side , which

is opposite to the smallest angle $/_b :.B=12$

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:

$A/sina = B/sinb=C/sinc ; A/sina=B/sinb$ or

$A/sin105=12/sin30 :. A= (12*sin105)/sin30~~ 23.18$ unit

Now we know sides $A=23.18 , B=12$ and their included angle

$/_c = 45^0$ . Area of the triangle is $A_t=(A*B*sinc)/2$

$:.A_t=(23.18*12*sin45)/2 ~~98.34$ sq.unit.

Area of the largest possible triangle is $98.34$ sq.unit [Ans]

A triangle has two corners with angles of ( pi ) / 4 ( pi ) / 4 and ( 7 pi )/ 12 ( 7 pi )/ 12 . If one side of the triangle has a length of 12 12 , what is the largest possible area of the triangle?