Using the law of cosines, write an algebraic proof to show that the angles in an equilateral triangle must equal 60°. Use "∧" to indicate exponents. For example, type a2 as a∧2. (Hint: Let s be a length of each side and x be the angle measure; then use these variables in the law of cosines.)
Answers
Cosine rule states that, if the sides of a triangle ABC are represented as a, b and c which are opposite to the angle A, B and C respectively, then,
a^2 = b^2 + c^2 - 2bc · cos A
b^2 = c^2 + a^2 - 2ca · cos B
c^2 = a^2 + b^2 - 2ab · cos C
Here we are given to prove that the angles must be equal to 60° in an equilateral triangle by using the law of cosines.
And we're given to let each side as 's' and each angle as 'x'.
So here,
a = b = c = s
and
⟨A = ⟨B = ⟨C = x
By using law of cosines, we get,
s^2 = s^2 + s^2 - 2 · s · s · cos x
=> s^2 = 2s^2 - 2s^2 · cos x
=> s^2 = 2s^2 (1 - cos x)
=> s^2 / 2s^2 = 1 - cos x
=> 1/2 = 1 - cos x
=> cos x = 1 - 1/2
=> cos x = 1/2
=> x = arccos(1/2)
=> x = (360n)° ± 60°
So we get that the each three angle measures (360n)° ± 60°, where n is for any whole number.
We know that the sum of the interior angles of any triangle is 180°. So, since we're given that each angle are represented as 'x', we can write,
3x = 180°
=> 3[(360n)° ± 60°] = 180°
=> (1080n)° ± 180° = 180°
On taking (1080n)° + 180° = 180°,
=> (1080n)° = 0
=> n = 0
So 'n' can be 0.
On taking (1080n)° - 180° = 180°,
=> (1080n)° = 360°
=> n = 360° / 1080°
=> n = 1/3
Here 'n' is not a whole number. So 'n' can't be 1/3.
So, taking n = 0, we get,
x = 60° OR x = - 60°
But x can't be -60° because it's a negative angle and a triangle can't have a negative angle.
So,
x = 60°
Hence Proved!