# How Many Acute Angles

Knowing that the big angle is 90, how many acute angles we have in this shape?

I know acute angle is less than 90, so we have 4 acute angles between the inner lines. Also we have 3 more acute angles combining the above angles. So the total will be 7 acute angles.

But the answer says its 9 acute angles, what am I missing?

• You forgot the two cases , where three angles are combined. – Peter Nov 9 '16 at 16:03
• I'd hardly call this a geometry problem any more than I'd call the beetles in four corners crawling toward each other a biology problem. This is a basic combinatorics problem that uses a picture of angles as it's subject. Anyway, you have 4 "distinct" acute angles. Combine two and you have 3 more. Combine three and you have 2 more. (Combine all four and you get a right angle). It's a basic question about the way to combine objects where only adjacent objects can be combined. The answer is to add 1 + 2 +3 +4 (but one if them is a right angle and doesn't count.) – fleablood Nov 9 '16 at 16:18
• In your case, we have 3 lines and the answer is 9 angles(see below). More interesting "How many acute angles we get if we would have used $n$ lines?", the answer is $\frac{(n+1)(n+2)}{2}-1$. – Math137 Nov 9 '16 at 16:21
• As always, I believe in expressing things simply when they are simple. I hope that unlike some other answers, mine makes the matter as simple as it really is. – Michael Hardy Nov 9 '16 at 16:21
• @MichaelHardy Your answer was very simple. But I think all the others were too. It's only the comments that were complicated. – fleablood Nov 9 '16 at 17:55

Call the five rays $A,B,C,D,E$, going clockwise, so that $A$ is horizontal and $E$ is vertical. Then acute angles correspond to $$AB,\quad AC,\quad AD, \quad BC,\quad BD,\quad BE,\quad CD, \quad CE, \quad DE$$ That's nine acute angles.
Let $O$ be the common vertex, and $A,B,C,D,E$ be the other 5 vertices in clockwise order. Observe that, we have the following 9 acute angles - $\angle AOB$, $\angle AOC$, $\angle AOD$, $\angle BOC$, $\angle BOD$, $\angle BOE$, $\angle COD$, $\angle COE$, $\angle DOE$.
There are ${5 \choose 2} - 1 = 9$ acute angles because you can select any two of the rays except the outermost pair, which are at right angles.