A triangle has two corners with angles of # pi / 2 # and # ( pi )/ 8 #. If one side of the triangle has a length of #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#