Two corners of a triangle have angles of $\frac{π}{3}$ $( pi )/ 3$ and $\frac{π}{4}$ $( pi ) / 4$ . If one side of the triangle has a length of $18$ $18$ , what is the longest possible perimeter of the triangle?

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Answer 1 · 2017-07-06T12:29:23

The perimeter is $= 64.7 u$ $=64.7u$

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Let

$ˆ A = \frac{1}{3} π$ $hatA=1/3pi$

$ˆ B = \frac{1}{4} π$ $hatB=1/4pi$

So,

$ˆ C = π - (\frac{1}{3} π + \frac{1}{4} π) = \frac{5}{12} π$ $hatC=pi-(1/3pi+1/4pi)=5/12pi$

The smallest angle of the triangle is $= \frac{1}{4} π$ $=1/4pi$

In order to get the longest perimeter, the side of length $18$ $18$

is $b = 18$ $b=18$

We apply the sine rule to the triangle $DeltaABC$

$a/sin hatA=c/sin hatC=b/sin hatB$

$a/sin (1/3pi) = c/ sin(5/12pi)=18/sin(1/4pi)=25.5$

$a=25.5*sin (1/3pi)=22.1$

$c=25.5*sin(5/12pi)=24.6$

The perimeter of triangle $DeltaABC$ is

$P=a+b+c=22.1+18+24.6=64.7$

Two corners of a triangle have angles of ( pi )/ 3 π3 ( pi )/ 3 and ( pi ) / 4 π4 ( pi ) / 4 . If one side of the triangle has a length of 18 18 18 , what is the longest possible perimeter of the triangle?