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Two corners of a triangle have angles of $\frac{π}{6}$ $pi / 6$ and $\frac{π}{2}$ $pi / 2$ . If one side of the triangle has a length of $3$ $3$ , what is the longest possible perimeter of the triangle?

1 Answer

Mar 7, 2017

$9 + 3 \sqrt{3}$ $9+3sqrt(3)$

Explanation:

The longest perimeter will occur if the given side length is the shortest side length i.e. if 3 is the length opposite the smallest angle, $\frac{π}{6}$ $pi/6$

enter image source here

By definition of $sin$ $sin$
$XXX \frac{3}{h} = sin (\frac{π}{6})$ $color(white)("XXX")3/h=sin(pi/6)$

$XXX \to h = \frac{3}{sin (\frac{π}{6})} = \frac{3}{\frac{1}{2}} = 6$ $color(white)("XXX")rarr h=3/sin(pi/6)=3/(1/2)=6$

Using the Pythagorean Theorem
$XXX x = \sqrt{6^{2} - 3^{2}} = \sqrt{27} = 3 \sqrt{3}$ $color(white)("XXX")x=sqrt(6^2-3^2)=sqrt(27)=3sqrt(3)$

Perimeter $= 3 + h + x = 3 + 6 + 3 \sqrt{3} = 9 + 3 \sqrt{3}$ $=3+h+x=3+6+3sqrt(3)=9+3sqrt(3)$

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