Problem: A farmer wants to fence a piece of his land. two sides of a triangular field have lengths of 120 ft. and 325 ft. the measure of the angle between those sides is 70°. How much fencing will be farmer need?
\large \bold{\blue{Solution:}}Solution: The triangular piece of land the farmer owned is an oblique triangle with the given of two sides and an included angle (SAS), the third side can be solved using the law of cosines:
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Answer:
Problem: A farmer wants to fence a piece of his land. two sides of a triangular field have lengths of 120 ft. and 325 ft. the measure of the angle between those sides is 70°. How much fencing will be farmer need?
\large \bold{\blue{Solution:}}Solution: The triangular piece of land the farmer owned is an oblique triangle with the given of two sides and an included angle (SAS), the third side can be solved using the law of cosines:
\sf x^2 = 120^2 + 325^2 - 2(120)(325)(cos \: 70 \degree)x
2
=120
2
+325
2
−2(120)(325)(cos70°)
\sf x^2 = 14,\!400 + 105,\!625 - 78,\!000(cos \: 70 \degree)x
2
=14,400+105,625−78,000(cos70°)
\sf x^2 = 14,\!400 + 105,\!625 - 26,\!677.57x
2
=14,400+105,625−26,677.57
\sf x^2 = 93,\!347.43x
2
=93,347.43
\sf \sqrt{x^2} = \sqrt{93,\!347.43}
x
2
=
93,347.43
\sf x = 305.53x=305.53
» So the measure of the third side is 305.53 feet. Find the perimeter of his triangular land by adding all the measure of the sides.
\sf 120ft + 325ft + 305.53ft = 750.53ft120ft+325ft+305.53ft=750.53ft
\begin{gathered} \large \therefore \underline{\boxed{\purple{ \begin{aligned} & \tt 750.53 \: feet \: of \\ & \tt fencing \: materials \end{aligned}}}} \end{gathered}
∴
750.53feetof
fencingmaterials
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