A triangle has two corners with angles of $\frac{π}{4}$ $pi / 4$ and $\frac{π}{6}$ $pi / 6$ . If one side of the triangle has a length of $12$ $12$ , what is the largest possible area of the triangle?

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Answer 1 · 2017-12-11T05:14:59

Largest possible area of the triangle is 98.3538

Given are the two angles $\frac{π}{4}$ $(pi)/4$ and $\frac{π}{6}$ $pi/6$ and the length 1

The remaining angle:

$= π - (\frac{π}{4}) + \frac{π}{6}) = \frac{7 π}{12}$ $= pi - ((pi)/4) + pi/6) = (7pi)/12$

I am assuming that length AB (12) 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 $= \frac{12^{2} \cdot sin (\frac{π}{4}) \cdot sin (\frac{7 π}{12})}{2 \cdot sin (\frac{π}{6})}$ $=( 12^2*sin(pi/4)*sin((7pi)/12))/(2*sin(pi/6))$

Area $= 98.3538$ $=98.3538$

A triangle has two corners with angles of pi / 4 π4 pi / 4 and pi / 6 π6 pi / 6 . If one side of the triangle has a length of 12 1212 , what is the largest possible area of the triangle?