3
$\begingroup$

enter image description here

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?

$\endgroup$
  • 3
    $\begingroup$ You forgot the two cases , where three angles are combined. $\endgroup$ – Peter Nov 9 '16 at 16:03
  • 1
    $\begingroup$ 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.) $\endgroup$ – fleablood Nov 9 '16 at 16:18
  • $\begingroup$ 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 $\frac{(n+1)(n+2)}{2}-1$. $\endgroup$ – Math137 Nov 9 '16 at 16:21
  • $\begingroup$ 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. $\endgroup$ – Michael Hardy Nov 9 '16 at 16:21
  • $\begingroup$ @MichaelHardy Your answer was very simple. But I think all the others were too. It's only the comments that were complicated. $\endgroup$ – fleablood Nov 9 '16 at 17:55
6
$\begingroup$

You have 9 different angles in that figure: enter image description here

$\endgroup$
  • $\begingroup$ @MichaelHardy, Anything else?! $\endgroup$ – Seyed Nov 9 '16 at 16:23
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 - $\angle AOB$, $\angle AOC$, $\angle AOD$, $\angle BOC$, $\angle BOD$, $\angle BOE$, $\angle COD$, $\angle COE$, $\angle DOE$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.