Two corners of a triangle have 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 $7$ $7$ , what is the longest possible perimeter of the triangle?

Question

1183 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-08T13:40:46

Largest possible area of the triangle is 21.2176

Given are the two angles $\frac{2 π}{3}$ $(2pi)/3$ and $\frac{π}{6}$ $pi/6$ and the length 7

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 (7) 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 $= \frac{7^{2} \cdot sin (\frac{π}{6}) \cdot sin (\frac{2 π}{3})}{2 \cdot sin (\frac{π}{6})}$ $=( 7^2*sin(pi/6)*sin((2pi)/3))/(2*sin(pi/6))$

Area $= 21.2176$ $=21.2176$

Two corners of a triangle have angles of 2π3 (2 pi )/ 3 and π6 ( pi ) / 6 . If one side of the triangle has a length of 7 7 , what is the longest possible perimeter of the triangle?