Two corners of a triangle have angles of $\frac{3 π}{8}$ $(3 pi ) / 8$ and $\frac{π}{12}$ $pi / 12$ . If one side of the triangle has a length of $9$ $9$ , what is the longest possible perimeter of the triangle?

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Answer 1 · 2017-07-07T12:15:46

The longest perimeter is $= 75.6 u$ $=75.6u$

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Let

$ˆ A = \frac{3}{8} π$ $hatA=3/8pi$

$ˆ B = \frac{1}{12} π$ $hatB=1/12pi$

So,

$ˆ C = π - (\frac{3}{8} π + \frac{1}{12} π) = \frac{13}{24} π$ $hatC=pi-(3/8pi+1/12pi)=13/24pi$

The smallest angle of the triangle is $= \frac{1}{12} π$ $=1/12pi$

In order to get the longest perimeter, the side of length $9$ $9$

is $b = 9$ $b=9$

We apply the sine rule to the triangle $DeltaABC$

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

$a/sin (3/8pi) = c/ sin(13/24pi)=9/sin(1/12pi)=34.8$

$a=34.8*sin (3/8pi)=32.1$

$c=34.8*sin(13/24pi)=34.5$

The perimeter of triangle $DeltaABC$ is

$P=a+b+c=32.1+9+34.5=75.6$

Two corners of a triangle have angles of (3 pi ) / 8 3π8(3 pi ) / 8 and pi / 12 π12 pi / 12 . If one side of the triangle has a length of 9 99 , what is the longest possible perimeter of the triangle?