A triangle has two corners with angles of $\frac{2 π}{3}$ $(2 pi ) / 3$ and $\frac{π}{6}$ $( pi )/ 6$ . If one side of the triangle has a length of $15$ $15$ , what is the largest possible area of the triangle?

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Answer 1 · 2017-11-13T11:29:25

Largest possible area of $Delta = 97.43$

The three angles are $(2pi)/3, (pi)/6 , pi - (2pi)/3 - pi/6 = pi/6$
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It’s an isosceles triangle.

$a / sin A = b / sin B = c / sin C$

$15 / sin (pi/6) = b / sin ((2pi)/3) = c / sin (pi/6)$

$:. c = 15$

$b = (15 * sin (2pi/3)) / sin (pi/6)$

$b = (15 * (sqrt3/2))/ (1/2) = 15 sqrt3$

Height $h = 15 * sin( pi /6) = 15/2$

Area of $Delta = (1/2) b h$

Area of $Delta = (1/2) * 15 sqrt3 * (15/2)$

$= (225 sqrt3) / 4 = 97.43$

A triangle has two corners with angles of (2 pi ) / 3 2π3 (2 pi ) / 3 and ( pi )/ 6 π6 ( pi )/ 6 . If one side of the triangle has a length of 15 1515 , what is the largest possible area of the triangle?