A triangle has two corners with angles of $\frac{π}{4}$ $( pi ) / 4$ and $\frac{7 π}{12}$ $( 7 pi )/ 12$ . If one side of the triangle has a length of $15$ $15$ , what is the largest possible area of the triangle?

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Answer 1 · 2017-12-08T14:50:08

Largest possible area of the triangle is 153.6779**

Given are the two angles $\frac{7 π}{12}$ $(7pi)/12$ and $\frac{π}{4}$ $pi/4$ and the length 15

The remaining angle:

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

I am assuming that length AB (15) 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 $= \frac{15^{2} \cdot sin (\frac{π}{4}) \cdot sin (\frac{7 π}{12})}{2 \cdot sin (\frac{π}{6})}$ $=( 15^2*sin(pi/4)*sin((7pi)/12))/(2*sin(pi/6))$

Area $= 153.6779$ $=153.6779$

A triangle has two corners with angles of π4 ( pi ) / 4 and 7π12 ( 7 pi )/ 12 . If one side of the triangle has a length of 1515 , what is the largest possible area of the triangle?