A triangle has two corners with angles of $\frac{π}{12}$ $pi / 12$ and $\frac{π}{12}$ $pi / 12$ . If one side of the triangle has a length of $4$ $4$ , what is the largest possible area of the triangle?

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Answer 1 · 2017-12-08T10:12:16

Largest possible area of the triangle = 4

Given are the two angles $\frac{π}{12}$ $(pi)/12$ and $\frac{π}{12}$ $pi/12$ and the length 4

The remaining angle:

$= π - ((\frac{π}{12}) + \frac{π}{12}) = \frac{5 π}{6}$ $= pi - (((pi)/12) + pi/12) =( 5pi)/6$

I am assuming that length AB (4) 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 $= \frac{4^{2} \cdot sin (\frac{π}{12}) \cdot sin (\frac{5 π}{6})}{2 \cdot sin (\frac{π}{12})}$ $=( 4^2*sin(pi/12)*sin((5pi)/6))/(2*sin(pi/12))$

Area $= 4$ $=4$

A triangle has two corners with angles of π12 pi / 12 and π12 pi / 12 . If one side of the triangle has a length of 44 , what is the largest possible area of the triangle?