Two corners of a triangle have angles of $\frac{7 π}{12}$ $(7 pi ) / 12$ and $\frac{π}{6}$ $pi / 6$ . If one side of the triangle has a length of $6$ $6$ , what is the longest possible perimeter of the triangle?

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Answer 1 · 2017-07-08T10:59:28

THe longest perimeter is $= 26.1 u$ $=26.1u$

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Let

$ˆ A = \frac{7}{12} π$ $hatA=7/12pi$

$ˆ B = \frac{1}{6} π$ $hatB=1/6pi$

So,

$ˆ C = π - (\frac{7}{12} π + \frac{1}{6} π) = \frac{1}{4} π$ $hatC=pi-(7/12pi+1/6pi)=1/4pi$

The smallest angle of the triangle is $= \frac{1}{6} π$ $=1/6pi$

In order to get the longest perimeter, the side of length $6$ $6$

is $b = 6$ $b=6$

We apply the sine rule to the triangle $DeltaABC$

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

$a/sin (7/12pi) = c/ sin(1/4pi)=6/sin(1/6pi)=12$

$a=12*sin (7/12pi)=11.6$

$c=12*sin(1/4pi)=8.5$

The perimeter of triangle $DeltaABC$ is

$P=a+b+c=11.6+6+8.5=26.1$

Two corners of a triangle have angles of (7 pi ) / 12 7π12(7 pi ) / 12 and pi / 6 π6 pi / 6 . If one side of the triangle has a length of 6 66 , what is the longest possible perimeter of the triangle?