A triangle has two corners with angles of $\frac{π}{4}$ $pi / 4$ and $\frac{π}{8}$ $pi / 8$ . If one side of the triangle has a length of $2$ $2$ , what is the largest possible area of the triangle?

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Answer 1 · 2017-12-11T08:06:59

Largest possible area of the triangle is 3.4142

Given are the two angles $\frac{π}{8}$ $(pi)/8$ and $\frac{π}{4}$ $pi/4$ and the length 2

The remaining angle:

$= π - (\frac{π}{8}) + \frac{π}{4}) = \frac{5 π}{8}$ $= pi - ((pi)/8) + pi/4) = (5pi)/8$

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 $= \frac{2^{2} \cdot sin (\frac{5 π}{8}) \cdot sin (\frac{π}{4})}{2 \cdot sin (\frac{π}{8})}$ $=( 2^2*sin((5pi)/8)*sin((pi)/4))/(2*sin(pi/8))$

Area $= 3.4142$ $=3.4142$

A triangle has two corners with angles of π4 pi / 4 and π8 pi / 8 . If one side of the triangle has a length of 22 , what is the largest possible area of the triangle?