Two corners of a triangle have angles of $\frac{π}{2}$ $( pi )/ 2$ and $\frac{π}{6}$ $( pi ) / 6$ . If one side of the triangle has a length of $4$ $4$ , what is the longest possible perimeter of the triangle?

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Two corners of a triangle have angles of $\frac{π}{2}$ $( pi )/ 2$ and $\frac{π}{6}$ $( pi ) / 6$ . If one side of the triangle has a length of $4$ $4$ , what is the longest possible perimeter of the triangle?

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Answer 1 · 2016-09-06T09:43:34

The longest possible perimeter is $18.9282$ $18.9282$

Explanation:

As two angles are $\frac{π}{2}$ $pi/2$ and $\frac{π}{6}$ $pi/6$ , third angle is

$π - \frac{π}{2} - \frac{π}{6} = \frac{6 π}{6} - \frac{3 π}{6} - \frac{π}{6} = \frac{2 π}{6} = \frac{π}{3}$ $pi-pi/2-pi/6=(6pi)/6-(3pi)/6-pi/6=(2pi)/6=pi/3$

Observe that as the angles are $30^{o}, 60^{o}, 90^{o}$ $30^o,60^o,90^o$ , the longest side (hypotenuse) is double the smallest side.

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The longest perimeter would be when the side $4$ $4$ would be opposite smallest angle $\frac{π}{6}$ $pi/6$ .

Then longest side of triangle is $2 \times 4 = 8$ $2xx4=8$ and third side using Pythagoras theorem is

$\sqrt{8^{2} - 4^{2}} = \sqrt{64 - 16} - \sqrt{48} = 6.9282$ $sqrt(8^2-4^2)=sqrt(64-16)-sqrt48=6.9282$

and longest possible perimeter is $4 + 8 + 6.9282 = 18.9282$ $4+8+6.9282=18.9282$

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Two corners of a triangle have angles of $\frac{π}{2}$ $( pi )/ 2$ and $\frac{π}{6}$ $( pi ) / 6$ . If one side of the triangle has a length of $4$ $4$ , what is the longest possible perimeter of the triangle?

1 Answer

Explanation:

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Two corners of a triangle have angles of ( pi )/ 2 π2 ( pi )/ 2 and ( pi ) / 6 π6 ( pi ) / 6 . If one side of the triangle has a length of 4 4 4 , what is the longest possible perimeter of the triangle?

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Explanation:

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Two corners of a triangle have angles of $\frac{π}{2}$ $( pi )/ 2$ and $\frac{π}{6}$ $( pi ) / 6$ . If one side of the triangle has a length of $4$ $4$ , what is the longest possible perimeter of the triangle?