Two angles of a triangle measure 30 and 45 degrees. If the side of the triangle opposite the 30-degree angle measures \(12\) units, what is the area of the triangle? Express your answer as a decimal to the nearest tenth.
+2668 Draw an altitude that forms both a 45-45-90 triangle and a 30-60-90 triangle
Because we have a 30-60-90 triangle, the base of the triangle is \({12 \over \sqrt3} = {4 \sqrt3}\)
We also have a 45-45-90 triangle. We know 1 of the legs is 12, meaning the other leg must also be 12.
Thus, the area of the triangle is \(12 \times (12 +4 \sqrt3) \div 2 \approx \color{brown}\boxed{113.6}\)
Here is a diagram:
+130071 Another way
The third angle = 105°
Using the Law of Sines the side opposite the 45° angle = x = can be found as
x / sin 45 = 12 /sin 30
x = 12 sin 45 / sin 30 = 12 sqrt (2)
So....using the area of a triangle knowing two sides and an included angle we have
A = (1/2) (12 sqrt 2 ) * ( 12 ) * [ sin (105°) ] =
Note : {sin 105 = sin 75 }
72sqrt (2) * sin (30 + 45) =
72 sqrt (2) [ sin 30 * cos 45 + sin 45 * cos 30 ] =
72 sqrt (2) [ sqrt 2 / 4 + sqrt 6 / 4 ] =
72 (2/4) + 18 sqrt (12) =
36 + 18 * 2 sqrt (3) =
36 + 36 sqrt (3) = 36 (1 + sqrt (3) ) units^2 ≈ 98.4 units^2
