A triangle has two corners with angles of $pi / 6$ and $(5 pi )/ 8$ . If one side of the triangle has a length of $8$ , what is the largest possible area of the triangle?

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Answer 1 · 2018-02-19T05:01:41

Area of triangle $A_t = (1/2) a b sin C ~~ color (red)(36$

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Given $hatA = pi/6, hat B (5pi)/8$ , one side = 8#

To find the largest possible area of the triangle.

Third angle $hatC = pi - pi/6 - (5pi)/8 = (5pi)/24$

To get the largest possible area, side8 should correspond to to the least angle.

$a/ sin ((5pi)/24) = b / sin ((5pi)/8) = c / sin (pi)/6$ using, law of sines.

$a = (8 * sin ((5pi)/24)) / sin (pi/6) = 9.7402$

$b = (8 * sin ((5pi)/8)) / sin (pi/6) = 14.7821$

Area of triangle $A_t = (1/2) a b sin C = (1/2) * 9.7402 * 14.7821 * sin (pi/6) ~~ color (red)(36$

A triangle has two corners with angles of pi / 6 pi / 6 and (5 pi )/ 8 (5 pi )/ 8 . If one side of the triangle has a length of 8 8 , what is the largest possible area of the triangle?