The sides of a triangle are m, n and \(\rm \sqrt{m^2+n^2+mn}\). What is the sum of the acute angles of the triangle?

Concept:

Cosine Rule of Triangle:

The square of the length of any side of a given triangle is equal to the sum of the squares of the length of the other sides minus twice the product of the other two sides multiplied by the cosine of angle included between them.

Consider, a, b, and c are lengths of the side of a triangle ABC as shown, then;

\(cos(x)=\frac{b^2+c^2-a^2}{2bc}\)

\(cos(y)=\frac{a^2+c^2-b^2}{2ac}\)

\(cos(z)=\frac{a^2+b^2-c^2}{2ab}\)

Calculation:

Let m = n = 1 unit

Then,

 \(\rm \sqrt{m^2+n^2+mn} \ \\\Rightarrow\rm \sqrt{1^2+1^2+1}= \sqrt 3\) 

Using cosine rule;

\(\rm \cos θ = {1^2+ 1^2 - {\sqrt 3}^2\over2\times 1 \times 1}\)

⇒ cos θ = -1/2

∴ θ = 120° 

Now, the sum of the acute angles of the triangle = 180° - 120° = 60°