So area (a) = 1/2 (base)(height) = 1/2bh
We have the b at 25m, but to find the h we need to drop an altitude or apothem down that is perpendicular to the base (90°). Now we have to do some trig to determine that h.
Since we're given that AB is 40m, and <B is 80°, we can set up a SOH-CAH-TOA equation:
The hypotenuse = 40m, and our height is opposite <B, therefore sin (80) = h/40,
h = 40 × sin (80) = 40 × 0.985 = 39.39m
Now, a = 1/2bh = 1/2 (25m)(39.4m)
[tex]a \: = 492.40 \: {m}^{2} [/tex]
