Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to solve the current problem

If O is the center of a circle with diameter 10 and the perimeter of AOB=16 then which is more x or 60

enter image description here

Now I know the triangle above is an isosceles triangle with 2 sides being 5 (since diameter is 10) and third side being 6 thus 2y+x=180. However I just cant figure out y or x.Are there any suggestions on how i could solve this problem.Without using Trig ratios or at least estimate whether it would be greater than 60?

share|cite|improve this question
You are I think calling $y$ the angle at the centre. Then it is $y+x+x=180^\circ$. For note that it is sides $OA$ and $OB$ that are equal, so $\angle OAB=\angle OBA=x$. The picture is kind of distorted! – André Nicolas Jul 10 '12 at 22:10
up vote 4 down vote accepted

The largest side of a triangle is opposite the largest angle. So $\angle AOB$ at the centre is bigger than either of the other two angles of the triangle. It follows that $\angle AOB$ is bigger than $60^\circ$, and $x$ is less than $60^\circ$.

If you wish to use more algebra, let $\angle AOB =w$. Then $w+x+x=180^\circ$. But $w\gt x$, so $x+x+x \lt 180^\circ$, and therefore $x \lt 60^\circ$.

share|cite|improve this answer

$\triangle AOB\;$ fits into an equilateral triangle (of perimeter $6+6+6=18$), therefore $ x <60^\circ$.

share|cite|improve this answer
What makes you say that triangle is equilateral. It cant ever be one. Look at the sides 5,5 and 6 ? Am i wrong ? – Rajeshwar Jul 10 '12 at 22:10
AOB is not itself equilateral, but it fits inside one. Specifically, construct D such that ADB is equilateral. Then O will lie inside AOB, and therefore angle $OAB$ is less than angle $DAB$ ... and the latter angle is 60°. – Henning Makholm Jul 10 '12 at 22:13
@Rajeshwar I just say, that yours fits in an equilateral triangle with perimeter $6+6+6=18$, which has all angles equal to $60^\circ$. So $x<60^\circ$. – draks ... Jul 10 '12 at 22:13
Thanks for clearing that up – Rajeshwar Jul 10 '12 at 22:29

First note that $OA = OB = 5$ and $AB = 6$. Hence, we have $AB > OA = OB$. Recall the sine rule that $$\dfrac{OA}{\sin(B)} = \dfrac{OB}{\sin(A)} = \dfrac{AB}{\sin(O)}$$ Since $\angle A = \angle B = x$, we have that $$\sin(x) = \sin(O) \times \dfrac56$$ Since $\sin$ is a strictly increasing function, we have that $\sin(x) < \sin(O) \implies x < \angle O$. We also have that $$x +x + \angle O = 180^{\circ} \implies 180^{\circ} > 3x \implies x < 60^{\circ}$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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