Question

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 degrees. Find the area of the park to the nearest square foot.

Answer 1

Answer:

The area of the park is 1,11,739 square feet.

Step-by-step explanation:

Since, the area of a triangle is,

$A=\frac{1}{2}\times s_1\times s_2\times sin\theta$

Where, $s_1$ and $s_2$ are the adjacent sides and $\theta$ is the included angle of these sides,

Here, the two adjacent sides of the park are 533 feet and 525 feet, while, the angle included by these sides is 53°.

That is,  $s_1$ = 533 ft, $s_2$ = 525 ft and $\theta$ = 53°,

Hence, the area of the park is,

$A=\frac{1}{2}\times 533\times 525\times sin 53^{\circ}$

$=\frac{279825\times 0.79863551004}{2}$

$=\frac{223478.181599}{2}=111739.090799\approx 111739\text{ square ft}$

Answer 2

Two adjacent sides of the park say x and y are,
x=533feet
y=525feet
A=53º
area=1/2*b*c*sin(A)
111739 feet^2