A triangle has two corners with angles of $\frac{3 π}{4}$ $(3 pi ) / 4$ and $\frac{π}{6}$ $( pi )/ 6$ . If one side of the triangle has a length of $7$ $7$ , what is the largest possible area of the triangle?

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Answer 1 · 2017-12-11T07:49:11

Largest possible area of the triangle is 33.4676

Given are the two angles $\frac{3 π}{4}$ $(3pi)/4$ and $\frac{π}{6}$ $pi/6$ and the length 7

The remaining angle:

$= π - (\frac{3 π}{4}) + \frac{π}{6}) = \frac{π}{12}$ $= pi - ((3pi)/4) + pi/6) = (pi)/12$

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 $= \frac{7^{2} \cdot sin (\frac{3 π}{4}) \cdot sin (\frac{π}{6})}{2 \cdot sin (\frac{π}{12})}$ $=( 7^2*sin((3pi)/4)*sin((pi)/6))/(2*sin(pi/12))$

Area $= 33.4676$ $= 33.4676$

A triangle has two corners with angles of 3π4 (3 pi ) / 4 and π6 ( pi )/ 6 . If one side of the triangle has a length of 77 , what is the largest possible area of the triangle?