# Math Help - Triangular Plot

1. ## Triangular Plot

A surveyor is mapping a triangular plot of land. He measures two of the sides and the angle formed by these two sides and finds that the lengths are 400 yards and 200 yards and the included angle is 50°.

What is the measure of the third side of the plot of land, to the
nearest yard?

What is the area of this plot of land, to the nearest square yard?

2. Hello, magentarita!

A surveyor is mapping a triangular plot of land.
Two of the sides are 400 and 200 yards and the included angle is 50°.

(a) What is the length of the third side of the triangle, to the nearest yard?
Law of Cosines . . .

$x^2 \;=\;400^2 + 200^2 - 2(400)(200)\cos50^o \;=\;97153.98245$

$x \;=\;311.695... \;\approx\;312$ yards

(b) What is the area of this plot of land, to the nearest square yard?
Formula: . $A \;=\;\tfrac{1}{2}\;\!ab\;\!\sin C$
. .
The area is one-half the product of two sides and the sine of the included angle.

Therefore: . $A \;=\;\tfrac{1}{2}(400)(200)\sin50^o \;=\;30641.777... \;\approx\;30,\!642$ square yeards.

3. ## ok............

Originally Posted by Soroban
Hello, magentarita!

Law of Cosines . . .

$x^2 \;=\;400^2 + 200^2 - 2(400)(200)\cos50^o \;=\;97153.98245$

$x \;=\;311.695... \;\approx\;312$ yards

Formula: . $A \;=\;\tfrac{1}{2}\;\!ab\;\!\sin C$
. . The area is one-half the product of two sides and the sine of the included angle.

Therefore: . $A \;=\;\tfrac{1}{2}(400)(200)\sin50^o \;=\;30641.777... \;\approx\;30,\!642$ square yeards.
Wonderfully done!