A park in a subdivision has a triangular shape. Two adjacent sides of the park are 533 feet and 525 feet. The angle between the sides is 53°. Find the area of the park to the nearest square foot.
Let $A$ and $B$ be points on the same side of the triangle such that $AB=533$. (We could choose the other side instead, but the answer would be the same.) Let $C$ be the foot of the altitude of the triangle from $B,$ and $D$ be the foot of the perpendicular from $C$ to $AB.$
[asy]
unitsize(0.2 cm);
pair A, B, C, D;
D = (0,0);
C = (54*19,0);
A = intersectionpoint(arc(C,533,0,180),arc(D,613,0,180));
B = (6*A + 4*C)/10;
draw(A--B--D);
draw(C--D);
label("$A$", A, dir(180));
label("$B$", B, dir(0));
label("$C$", C, dir(0));
label("$D$", D, dir(180));
label("$525$", (B + C)/2, dir(0));
label("$533$", (A + B)/2, dir(180));
[/asy]
Then $AC = 525 \cos 53^\circ = 525( \tfrac{1}{2}) = \tfrac{525}{2}$ and $BD = 525 \sin 53^\circ = 525( \tfrac{\sqrt{3}}{2}) = \tfrac{525 \sqrt{3}}{2},$ so
\[[ABC] = \frac{1}{2} \cdot \frac{525}{2} \cdot \frac{525 \sqrt{3}}{2} = \frac{525^2 \sqrt{3}}{8} \approx \boxed{55207}.\]
show the work simply and without explanation
We can use the formula for the area of a triangle given two sides and the included angle:
\[[ABC] = \frac{1}{2} \cdot AB \cdot AC \cdot \sin \angle B\]
\[[ABC] = \frac{1}{2} \cdot 533 \cdot \frac{525}{2} \cdot \sin 53^\circ\]
\[[ABC] = \frac{1}{2} \cdot 533 \cdot \frac{525}{2} \cdot \frac{\sqrt{3}}{2}\]
\[[ABC] \approx \boxed{55207}\]