A triangle has two corners with angles of $\frac{π}{2}$ $pi / 2$ and $\frac{3 π}{8}$ $(3 pi )/ 8$ . If one side of the triangle has a length of $9$ $9$ , what is the largest possible area of the triangle?

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Answer 1 · 2017-12-08T14:17:05

Largest possible area of the triangle is 97.7756

Given are the two angles $\frac{π}{2}$ $(pi)/2$ and $\frac{3 π}{8}$ $(3pi)/8$ and the length 9

The remaining angle:

$= π - (\frac{π}{2}) + \frac{3 π}{8}) = \frac{π}{8}$ $= pi - ((pi)/2) + (3pi)/8) = (pi)/8$

I am assuming that length AB (9) 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 $= \frac{9^{2} \cdot sin (\frac{π}{2}) \cdot sin (\frac{3 π}{8})}{2 \cdot sin (\frac{π}{8})}$ $=( 9^2*sin(pi/2)*sin((3pi)/8))/(2*sin(pi/8))$

Area $= 97.7756$ $=97.7756$

A triangle has two corners with angles of π2 pi / 2 and 3π8 (3 pi )/ 8 . If one side of the triangle has a length of 99 , what is the largest possible area of the triangle?