The algebraic proof shows that the angles in an equilateral triangle must equal 60° each
<h3>Laws of cosines </h3>
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: brainly.com/question/2866347
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