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?

Question

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?

Grace Ashcraft · Accepted Answer

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

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.

Grace Ashcraft · Answer

undefined To find the area of the triangle, you can use Heron's formula, which requires knowing the lengths of all three sides. However, in this case, we only have the length of side B, which is 17. Without the lengths of the other sides, we can't directly apply Heron's formula.

To proceed, we need more information about the triangle, such as the lengths of the other sides or additional angles. Without this information, we cannot determine the area of the triangle.