Deteriming an angle without Trig. ratios

Question

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?

Answer 1

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 $x \lt 60^\circ$.

Answer 2

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

Answer 3

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