A triangle has sides of integer lengths 3, 6 and x. For how many values of x will the triangle be acute?

\(\text{The triangle will be acute for }\\ 3 < x < \sqrt{3^2 + 6^2} = \\ 3 < x < 3\sqrt{5}\\ \text{That's a continuum of values so asking how many there are is sort of meaningless}\)

Actually, the answer key says it is 1. Do you know how?

oh it says integer lengths... DOH

\(3 < x < 3 \sqrt{5} \approx 6.7\\ \text{That gives you 4, 5, and 6}\)

4 gives you an angle of about 36.36 degrees

5 gives you an angle of about 56.25 degrees

6 gives you an angle of about 75.53 degrees

all 3 are acute triangles