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 $1$ $1$ , what is the largest possible area of the triangle?

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Answer 1 · 2017-12-08T15:02:27

Largest possible area of the triangle is 0.433

Given are the two angles $\frac{2 π}{3}$ $(2pi)/3$ and $\frac{π}{6}$ $pi/6$ and the length 1

The remaining angle:

$= π - (\frac{2 π}{3}) + \frac{π}{6}) = \frac{π}{6}$ $= pi - ((2pi)/3) + pi/6) = (pi)/6$

I am assuming that length AB (1) is opposite the smallest angle.

Using the ASA

Area $= \frac{c^{2} \cdot sin (A) \cdot sin (B)}{2 \cdot sin (C)}$ $=(c^2*sin(A)*sin(B))/(2*sin(C)$

Area $= \frac{1^{2} \cdot sin (\frac{π}{6}) \cdot sin (\frac{2 π}{3})}{2 \cdot sin (\frac{π}{6})}$ $=( 1^2*sin(pi/6)*sin((2pi)/3))/(2*sin(pi/6))$

Area $= 0.433$ $=0.433$

A triangle has two corners with angles of 2π3 (2 pi ) / 3 and π6 ( pi )/ 6 . If one side of the triangle has a length of 11 , what is the largest possible area of the triangle?