A triangle has sides A, B, and C. If the angle between sides A and B is #(pi)/6#, the angle between sides B and C is #(5pi)/12#, and the length of B is 17, what is the area of the triangle?

1 Answer
Jun 4, 2016

It turns out to be simply #{289}/4=72.25# square units.

Explanation:

The three angles of the triangle add up to #\pi# radians, so the angle between #A# and #C# is #\pi-(\pi/6)-({5\pi}/12)={5\pi}/12# radians.

Then the angles on both ends of side #C# are congruent and thus the triangle is isosceles, with side #A# congruent with side #B#. So side #A# measures #17# units along with side #B#. Now the area of the triangle is half the product of the two sides #A# and #B# times the sine of the angle between them:

#sin (\pi/6)=1/2#

Area = #(1/2)xx(17)xx(17)xx(1/2)={289}/4# square units.