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The algebraic proof shows that the angles in an equilateral triangle must equal 60° each
Laws of cosines
From the question, we are to use the law of cosines to write an algebraic proof that shows that the angles in an equilateral triangle must equal 60°.
Given any triangle ABC, the measures of angles A, B, and C by the law of cosines are
cos A = (b^2 + c^2 - a^2)/2bc
cos B= (a^2 + c^2 - b^2)/2ac
cos C = (a^2 + b^2 - c^2)/2ab
Now, given that the triangle is equilateral, with each of the side lengths equal to s
That is, a = b = c = s
Then, we can write that
cos A = (s^2 + s^2 - s^2)/(2s×s)
cos A = (s^2 )/(2s^2)
cos A = 1/2
cos A = 0.5
∴ A = cos⁻¹(0.5)
A = 60°
Also
cos B = (s^2 + s^2 - s^2)/(2s×s)
cos B = (s^2 )/(2s^2)
cos B = 1/2
cos B = 0.5
∴ B = cos⁻¹(0.5)
B = 60°
and
cos C = (s^2 + s^2 - s^2)/(2s×s)
cos C = (s^2 )/(2s^2)
cos C = 1/2
cos C = 0.5
∴ C = cos⁻¹(0.5)
C = 60°
Thus,
A = 60°, B = 60° and C = 60°
Hence, the algebraic proof above shows that the angles in an equilateral triangle must equal 60° each.
Learn more on The law of cosines here: https://brainly.com/question/2866347
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