# Deteriming an angle without Trig. ratios

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

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?

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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

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 $x \lt 60^\circ$.

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

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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 $$x +x + \angle O = 180^{\circ} \implies 180^{\circ} > 3x \implies x < 60^{\circ}$$

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