A triangle has two corners with angles of $\frac{2 π}{3}$ $(2 pi ) / 3$ and $\frac{π}{4}$ $( pi )/ 4$ . 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 · 2017-12-08T09:53:16

Largest possible area of the triangle = 4.7321

Given are the two angles $\frac{2 π}{3}$ $(2pi)/3$ or $120^{\circ}$ $120^@$ and $\frac{π}{4}$ $pi/4$ or $45^{\circ}$ $45^@$ and the length 12

The remaining angle:

$180^{\circ} - (120^{\circ} + 45^{\circ}) = 15^{\circ}$ $180^@-(120^@+45^@)=15^@$

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 $= \frac{2^{2} \cdot sin (45) \cdot sin (120)}{2 \cdot sin (15)}$ $=( 2^2*sin(45)*sin(120))/(2*sin(15))$

Area $= 4.7321$ $=4.7321$

A triangle has two corners with angles of 2π3 (2 pi ) / 3 and π4 ( pi )/ 4 . If one side of the triangle has a length of 1212 , what is the largest possible area of the triangle?