A triangle has two corners with angles of $\frac{π}{2}$ $pi / 2$ and $\frac{π}{8}$ $( pi )/ 8$ . 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 · 2018-07-27T08:04:04

$81/2(\sqrt2+1)=97.776\ \text{unit}^2$

Let the angles of a $\triangle ABC$ are $A=\pi/2$ , $B=\pi/8$ hence third angle $C$ is

$C=\pi-\pi/2-\pi/8$

$={3\pi}/8$

The area of given triangle will be largest only when the given side say $b=9$ is opposite to the smallest angle $B={pi}/8$

Now, using sine rule in given triangle as follows

$\frac{b}{\sin B}=\frac{c}{\sin C}$

$\frac{9}{\sin (\pi/8)}=\frac{c}{\sin ({3\pi}/8)}$

$c=\frac{9\sin({3\pi}/8)}{\sin(\pi/8)}$

$=9(\sqrt2+1)$

Hence, the largest area of given right triangle with legs $9$ & $9(\sqrt2+1)$ is given as

$=1/2(9)(9(\sqrt2+1))$

$=81/2(\sqrt2+1)$

$=97.776\ \text{unit}^2$

A triangle has two corners with angles of pi / 2 π2 pi / 2 and ( pi )/ 8 π8 ( pi )/ 8 . If one side of the triangle has a length of 9 99 , what is the largest possible area of the triangle?