help and give explanation
How many acute angles are formed by the 4 intersecting lines shown?
[asy]
import olympiad;
size(150);
pair A=(0,0), B = 4*dir(0), C=4*dir(40),D=4*dir(100),E=4*dir(150),F=4*dir(180),G=4*dir(220),H=4*dir(280),I=4*dir(330);
dot(A,red);
draw(A--B,EndArrow);
draw(anglemark(B,A,C,20),black);
label("$40^\circ$",anglemark(B,A,C),4*dir(20));
draw(A--C,EndArrow);
draw(A--D,EndArrow);
draw(A--E,EndArrow);
draw(anglemark(D,A,E,20),black);
label("$50^\circ$",anglemark(D,A,E),4*dir(120));
draw(A--F,EndArrow);
draw(A--G,EndArrow);
draw(A--H,EndArrow);
draw(anglemark(G,A,H,20),black);
label("$60^\circ$",anglemark(G,A,H),4*dir(250));
draw(A--I,EndArrow);
[/asy]
Here's the Asymptote code in picture form:
We can use vertical angles to fill in three of the missing angles. We have:
Since the total sum of angles in a full rotation is 360 degrees, the angles must sum up to 360. The two missing angles must be vertical angles too, so they are the same. Let us make the missing angles \(x\), and so we have:
\(2x = 360 - (2\cdot60-2\cdot40-2\cdot50)\)
\(x=50\)
Every angle is acute, so we already have 8 acute angles that we know of.
We can also combine two angles to make acute angles. The combinations below work:
30, 40
30, 50
40, 30
50, 30
Now we know 4 combinations of two angles that work.
No combinations of three angles will work because the smallest one is already obtuse.
So, the answer is 12 angles.
