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Old November 19th, 2008, 08:15 PM
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Default 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?


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Old November 19th, 2008, 11:46 PM
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Hello, magentarita!

Quote:
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




Quote:
(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.

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Old November 20th, 2008, 07:07 AM
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Default ok............

Quote:
Originally Posted by Soroban View Post
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!
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