A triangle has two corners with angles of $\frac{2 π}{3}$ $(2 pi ) / 3$ and $\frac{π}{4}$ $( pi )/ 4$ . If one side of the triangle has a length of $2$ $2$ , what is the largest possible area of the triangle?

Question

1205 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-13T12:12:55

Area of the largest possible triangle is $15.08$ $15.08$ sq.unit.

Angle between Sides $A and B$ $A and B$ is $∠ c = \frac{2 π}{3} = 120^{0}$ $/_c= (2pi)/3=120^0$

Angle between Sides $B and C$ $B and C$ is $/_a= pi/4=45^0 :.$

Angle between Sides $C and A$ is $/_b= 180-(120+45)=15^0$

For largest area of triangle $2$ should be smallest side , which

is opposite to the smallest angle $(/_b=15^0)$ , i.e $B=2$

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:

$A/sina = B/sinb=C/sinc ; B=2 :. A/sina=B/sinb$ or

$A/sin45=2/sin15 :. A= 2* sin45/sin15~~ 5.46(2dp)$

Now we know sides $A=5.46 , B=2.0$ and their included angle

$/_c = 120^0$ . Area of the triangle is $A_t=(A*B*sinc)/2$

$:.A_t=(5.46*cancel2*sin 120)/cancel2 ~~ 4.73$ sq.unit.

Area of the largest possible triangle is $15.08$ sq.unit [Ans]

A triangle has two corners with angles of (2 pi ) / 3 2π3 (2 pi ) / 3 and ( pi )/ 4 π4 ( pi )/ 4 . If one side of the triangle has a length of 2 22 , what is the largest possible area of the triangle?