A triangle has two corners with angles of $\frac{π}{3}$ $( pi ) / 3$ and $\frac{π}{6}$ $( pi )/ 6$ . If one side of the triangle has a length of $14$ $14$ , what is the largest possible area of the triangle?

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Answer 1 · 2017-12-11T06:16:09

Largest possible area of the triangle is 169.741

Given are the two angles $(\frac{π}{3})$ $(pi/3)$ and $\frac{π}{6}$ $pi/6$ and the length 14

The remaining angle:

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

I am assuming that length AB (14) 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 $= \frac{14^{2} \cdot sin (\frac{π}{2}) \cdot sin (\frac{π}{3})}{2 \cdot sin (\frac{π}{6})}$ $=( 14^2*sin((pi)/2)*sin((pi)/3))/(2*sin(pi/6))$

Area $= 169.741$ $=169.741$

A triangle has two corners with angles of π3 ( pi ) / 3 and π6 ( pi )/ 6 . If one side of the triangle has a length of 1414 , what is the largest possible area of the triangle?