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 $6$ $6$ , what is the largest possible area of the triangle?

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Answer 1 · 2017-12-08T13:26:44

Largest possible area of the triangle is 15.5885

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

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

Area $= 15.5885$ $=15.5885$

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 66 , what is the largest possible area of the triangle?