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 $3$ $3$ , what is the largest possible area of the triangle?

Question

1222 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-08T13:29:53

Largest possible area of the triangle is 3.8971

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

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

Area $= 3.8971$ $=3.8971$

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