The straight lines x+y-4=0, 3x+y-4=0 and x+3y-4=0 form a triangle which is:

The given lines are x+y-4=0, 3x+y-4=0 and x+3y-4=0.

First, let's find the slopes of these lines.

The slope of a line given by the equation ax+by+c=0 is -a/b.

So, the slopes of the given lines are:

1. For x+y-4=0, slope = -1/1 = -1
2. For 3x+y-4=0, slope = -3/1 = -3
3. For x+3y-4=0, slope = -1/3

Now, let's find the angles between these lines using the formula for the angle θ between two lines with slopes m1 and m2:

tan(θ) = |(m1 - m2) / (1 + m1*m2)|

1. Angle between lines 1 and 2: tan(θ) = |(-1 - (-3)) / (1 + -1*-3)| = 2/4 = 0.5, so θ = arctan(0.5) ≈ 26.57°
2. Angle between lines 2 and 3: tan(θ) = |(-3 - (-1/3)) / (1 + -3*-1/3)| = 2.67/2 = 1.335, so θ = arctan(1.335) ≈ 53.13°
3. Angle between lines 1 and 3: tan(θ) = |(-1 - (-1/3)) / (1 + -1*-1/3)| = 0.67/2 = 0.335, so θ = arctan(0.335) ≈ 18.43°

Since all angles are less than 90°, the triangle formed by these lines is acute.