A triangle has two corners with angles of $\frac{π}{4}$ $pi / 4$ and $\frac{π}{6}$ $pi / 6$ . If one side of the triangle has a length of $3$ $3$ , what is the largest possible area of the triangle?

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Answer 1 · 2018-07-15T12:15:05

Largest possible area of triangle is $6.15$ $6.15$ sq.unit

Angle between Sides $A and B$ $A and B$ is $∠ c = \frac{π}{4} = 45^{0}$ $/_c= pi/4=45^0$

Angle between Sides $B and C$ $B and C$ is $/_a= pi/6=30^0 :.$

Angle between Sides $C and A$ is $/_b= 180-(45+30)=105^0$

Length of one side is $3$ , For largest area of triangle $3$ should

be smallest side , which is opposite to the smallest angle

$/_a=30^0 :. A=3$ 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: $A/sin a = B/sin b=C/sin c ; A=3$

$:. A/sin a=B/sin b or 3/sin 30 = B/sin 105$ or

$B= 3* sin 105/sin 30 ~~ 5.8$ Now we know sides

$A=3 , B=5.8$ and their included angle $/_c =45^0$ .

Area of the triangle is $A_t=(A*B*sin c )/2$

$:. A_t=(3*5.8*sin 45 )/2 ~~ 6.15$ sq.unit

Largest possible area of triangle is $6.15$ sq.unit [Ans]

A triangle has two corners with angles of pi / 4 π4 pi / 4 and pi / 6 π6 pi / 6 . If one side of the triangle has a length of 3 33 , what is the largest possible area of the triangle?