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

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Answer 1 · 2017-12-23T13:00:48

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

Angle between Sides $A and B$ $A and B$ is

$∠ c = \frac{3 π}{4} = \frac{3 \cdot 180}{4} = 135^{0}$ $/_c= (3pi)/4=(3*180)/4=135^0$

Angle between Sides $B and C$ $B and C$ 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]

A triangle has two corners with angles of (3 pi ) / 4 3π4 (3 pi ) / 4 and ( pi )/ 8 π8 ( pi )/ 8 . If one side of the triangle has a length of 12 1212 , what is the largest possible area of the triangle?