# A farmer plans to build a triangular fence with side lengths of 500 m, 461 m, and 408 m. Determine the measures of the angles?

### 7 Answers

- 1 month ago
Let the lengths of the three sides of the triangle be:

a = 500, b = 461 and c = 408

The angle A (opposite to side a) may be obtained using the cosine rule as follows

a^2 = b^2 + c^2 - 2 b c cos (A)

500^2 = 461^2 + 408^2 - 2 (461)(408) cos (A)

Solve for cos (A)

cos(A) = -128985 / -376176 = 0.34288471353

A = arccos(0.34288471353) = 69.9472758 degrees

The angle B (opposite to side b) may be obtained using the cosine rule as follows

b^2 = a^2 + c^2 - 2 a c cos (B)

461^2 = 500^2 + 408^2 - 2 (500)(408) cos (B)

cos(B) = -203943 / -408000 = 0.49986029411

B = arccos(0.49986029411) = 60.0092424 dgrees

angle C may be found using the fact that the sum of all angles in a triangle is 180 degrees

A + B + C = 180

C = 180 - 69.9472758 - 60.0092424 = 50.0434818 degrees

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- Jeff AaronLv 71 month ago
Cosine Rule:

c^2 = a^2 + b^2 - 2ab*cos(C)

500^2 = 461^2 + 408^2 - 2*461*408*cos(C)

250000 = 212521 + 166464 - 376176*cos(C)

376176*cos(C) = 212521 + 166464 - 250000

376176*cos(C) = 128985

cos(C) = 128985 /376176

cos(C) = 42995/125392

Since C must be between 0 and pi radians (0° to 180°):

C = arccos(42995/125392)

C =~ 1.22 radians (69.95°)

Use the same method for the other two angles.

- KrishnamurthyLv 71 month ago
A farmer plans to build a triangular fence

with side lengths of 500 m, 461 m, and 408 m. Determine the measures of the angles?

A: 69.95°

B: 60.01°

C: 50.04°

Source(s): http://www.trianglecalculator.net/ - Anonymous1 month ago
I do not help people appear smarter than they actually are.

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- SlowfingerLv 61 month ago
Answer:

A=70°, B=60°, C=50°

Law of Cosines

c² = a² + b² - 2ab cos C

where C denotes the angle contained between sides of lengths a and b.

Knowing all sides, the angle between a and b is

cos C = (a² + b² - c²) / (2ab)

C = arccos [(a² + b² - c²) / (2ab)]

Let

a=500

b=461

c=408

and angles A, B and C are opposite to sides a, b and c, respectively.

C = arccos [(500² + 461² - 408²) / (2*500*461)] ≈ 50°

Law of Sines

c / sin C = b / sin B

B = arcsin ((b/c) * sin C)

B = arcsin ((461/408) * sin 50°) ≈ 60°

Naturally, A = 180°- B - C = 180°- 60° - 50° = 70°

- 1 month ago
Law of cosines:

500^2 = 461^2 + 408^2 - 2 * 461 * 408 * cos(C)

461^2 = 500^2 + 408^2 - 2 * 500 * 408 * cos(B)

408^2 = 500^2 + 461^2 - 2 * 500 * 461 * cos(A)

500^2 - (461^2 + 408^2) = -2 * 461 * 408 * cos(C)

461^2 + 408^2 - 500^2 = 2 * 461 * 408 * cos(C)

(461^2 + 408^2 - 500^2) / (2 * 461 * 408) = cos(C)

arccos((461^2 + 408^2 - 500^2) / (2 * 461 * 408)) = C

arccos((500^2 + 408^2 - 461^2) / (2 * 500 * 408)) = B

arccos((500^2 + 461^2 - 408^2) / (2 * 500 * 461)) = A

Make sure your calculator is in degree mode

Make sure you use brackets appropriately.

arccos(t) is also cos-1(t) on your calculator.