Two corners of a triangle have angles of $pi / 8$ and $pi / 6$ . If one side of the triangle has a length of $7$ , what is the longest possible perimeter of the triangle?

Question

1223 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-11T06:07:35

Longest possible perimeter of the triangle is 31.0412

Given are the two angles $(pi)/6$ and $(pi)/8$ and the length 1

The remaining angle:

$= pi - (((pi)/6) + (p)/8) = (17pi)/24$

I am assuming that length AB (7) is opposite the smallest angle

$a / sin A = b / sin B = c / sin C$

$7 / sin ((pi)/6) = b / sin ((pi) /8) = c / ((17pi) / 24)$

$b = (7*sin((3pi)/8)) / sin ((pi) /6) = 12.9343$

$c = (7*sin ((17pi)/24)) / sin ((pi)/6) = 11.1069$

Longest possible perimeter of the triangle is = $(a+b+c) = (7+12.9343+11.1069) = 31.0412$

Two corners of a triangle have angles of pi / 8 pi / 8 and pi / 6 pi / 6 . If one side of the triangle has a length of 7 7 , what is the longest possible perimeter of the triangle?