How can we determine the mass of the largest angle of an obtuse triangle with side lengths as three consecutive natural numbers?

Answer 1

Very interesting!

Let the numbers be a (a+1) and (a+2)

By the cosine law

(a+2)² = a² + (a+1)² - 2 a (a+1) cosθ

From this you get

cosθ = (a - 3)/ 2a

Now, cos has to be <0 (obtuse angle)

Solving this inequality, you will see that a should be exactly less then 3

Meaning a could be either 1 or 2

When a=1 we will get triangle (1,2,3) - which is not a triangle (1+2 has to be > 3)

So the only solution is when a=2

So we get a triangle (2,3,4)

cosθ = -1/4

And

θ = acos(-1/4) = 104.47^°

^{You can verify next triangle is (3,4,5) - which is the right triangle, so} θ=90

All following triangles will be acute!

Answer 2

I assume you mean the "measure".

Let the 3 natural numbers be n-1, n and n+1

then n-1 + n + n+1 = 3n = 180

and then numbers are 59, 60 and 61