SOLUTION: A triangular city lot bounded by 3 streets has a length of 300 fetter on one street, 250 feet on the second, and 420 feet on the third. What is the measure of the largest angle for
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Question 1030006: A triangular city lot bounded by 3 streets has a length of 300 fetter on one street, 250 feet on the second, and 420 feet on the third. What is the measure of the largest angle formed by these streets?
Found 2 solutions by stanbon, addingup:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
: A triangular city lot bounded by 3 streets has a length of 300 fetter on one street, 250 feet on the second, and 420 feet on the third. What is the measure of the largest angle formed by these streets?
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Largest angle is the angle opposite the longest side.
Use the Law of Cosines to find the angle.
Cheers,
Stan H.
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Answer by addingup(3677) (Show Source): You can put this solution on YOUR website!
When we know the three sides of a triangle, we call it an SSS triangle (we know the Side-Side-Side).
To find the angles of an SSS triangle we apply the Law of Cosines to find two of the angles, and the third simply by addition and subtraction, like this:
Let the angles be A, B, and C and the sides a, b, c. And:
a = 300
b = 250
c = 420
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cos C = (a^2+b^2-c^2)/2ab
cos A = (b^2+c^2-a^2)/2bc
cos B = (c^2+a^2-b^2)/2ca
NOTE: They are the same formula, the sides move around depending on which angle you are trying to find
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Finding angle C:
Cos C = (300^2+250^2-420^2)/2(300*250)
Cos C = (90,000+62,500-176,400)/150,000
Cos C = -23,900/150,000
Cos C = -0.159
C = cos^-1(0.159) Use the Cos^-1 function in your calculator
C = 80.85 degrees
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Find angle A:
Cos A = (250^2+420^2-300^2+)/2(300*250)/2(250*420)
Cos A = (62,500+176,400-90,000)/210,000
Cos A = -0.71
A = cos^-1(0.71)= 44.765 degrees
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Find angle B:
B = 180-(80.85+44.765)= 54.385
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