Question

The law of cosines is a2+b2 - 2abcosC = c^2 find the value of 2abcosC .... A. 40 B. -40 C. 37 D. 20

(The sides are 2,4, and 5; A to B is 2, B to C is 4, and A to C is 5.)

Answer 1

a² + b² - 2abcosC = c²

Аngle C lies opposite to the side AB, so "c" in the formula it is AB in your triangle

4² + 5² - 2abcosC = 2²
16 + 25 - 2abcosC = 4
41 - 4 = 2abcosC
2abcosC = 37

Answer 2

Given the Law of Cosines, the value of 2abcosC after plugging in the values of a, b, and c is: $\mathbf{2abcosC = 37}$

Law of Cosines is given as: $a^2+b^2 - 2abcosC = c^2$

Given also are the sides of a triangle:

• a = 4 (side B to C)
• b = 5 (side A to C)
• c = 2 (side A to B)

Plug in the values into the Law of Cosines,  $a^2+b^2 - 2abcosC = c^2$ to find $\mathbf{2abcosC}$

• Thus:

$4^2+5^2 - 2abcosC = 2^2\\\\16 + 25 - 2abcosC = 4\\\\41 - 2abcosC = 4$

• Subtract 41 from both sides

$41 - 2abcosC - 41 = 4-41\\\\-2abcosC = -37$

• Divide both sides by -1

$\mathbf{2abcosC = 37}$

Therefore, given the Law of Cosines, the value of 2abcosC after plugging in the values of a, b, and c is: $\mathbf{2abcosC = 37}$

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