# Circle theorem/triange angle question

I am doing practise papers and there is one question I cannot understand even with the mark scheme. I have added the pictures below:

Mark scheme:

The question is to find angle x (23 degrees, apparently).

At first I thought that angle OCD might be 69 degrees because of the alternate angle theorem, but I realised that it wasn't on a tangent so that wouldn't work. Other than that I have no idea - could someone please explain (in detail if possible, maths isn't my strength!) the steps to find it? Thanks in advance!

• <D = <DCO [radii of the same circle] = 2x [exterior angle of triangle]. 69 degrees = 2x + x [exterior angle of triangle]. – Mick May 31 '14 at 11:05
• OB and BE cannot be equal. – preferred_anon May 31 '14 at 11:06

The triangle $\;\Delta OCD\;$ is isosceles (why?), and thus $\;\angle OCD=\angle ODC\;$, and since $\;\angle DOC\;$ is the other angle , it must be that
$$111^\circ-x^\circ=\angle DOC=180^\circ-2\angle OCD$$
But $\;\angle OCD\;$ is an exterior one to $\;\Delta OCE\;$ and thus $\;\angle OCD=2x^\circ\;$ (why?), so we get
$$111^\circ-x^\circ=180-4x^\circ\implies3x^\circ=69^\circ\implies x=23^\circ$$
• @user135285, well $\;\angle OCD=2x^\circ\;$ because it is an exterior angle to $\;\Delta OCE\;$ and thus equal to the sum of the two angles of this triangle not adjacent to it... – DonAntonio May 31 '14 at 11:21