Question

apply: circle problem #4
point o is the centre of the circle.
what is the value of $\angle x^\circ$?
(there is a circle with center o, points a, b, c on the circumference. angle aob is 160 degrees, and we need to find angle acb which is x degrees.)

Explanation:

Step1: Recall the property of triangles in a circle

In triangle \(OBC\) and \(OAC\), \(OB = OC = OA\) (radii of the circle), so triangle \(ABC\) is isosceles? Wait, no, actually, we can use the property that the sum of angles in a triangle is \(180^\circ\), and also, the central angle is \(160^\circ\), so the triangle \(OBC\) and \(OAC\) are isosceles with \(OB = OC\) and \(OA = OC\). Wait, more accurately, in triangle \(ABC\)? No, let's think again. The central angle \(\angle BOC\) is \(160^\circ\)? Wait, no, the angle at \(O\) is \(160^\circ\), so \(\angle AOB = 160^\circ\)? Wait, the diagram shows points \(B\), \(O\), \(A\) with angle \(160^\circ\) at \(O\), and point \(C\) on the circumference. So triangle \(ABC\) with \(O\) as center. So \(OB = OC = OA\) (radii). So triangle \(OBC\) and \(OAC\) are isosceles. Wait, actually, the triangle \(ABC\) is an isosceles triangle? No, let's use the property of the inscribed angle and central angle. Wait, no, here we have a triangle with two sides as radii? Wait, no, the angle at \(C\) is \(x\), and we need to find \(x\). Let's consider triangle \(OBC\) and \(OAC\). Wait, \(OB = OC\) and \(OA = OC\), so \(\angle OBC=\angle OCB\) and \(\angle OAC=\angle OCA\). But maybe a better approach: the sum of angles in a triangle is \(180^\circ\), and the central angle is \(160^\circ\), so the remaining angle at \(O\) for the triangle? Wait, no, the triangle \(ABC\) has \(O\) as the center, so \(OB = OC = OA\), so triangle \(OBC\) and \(OAC\) are isosceles. Wait, actually, the angle at \(O\) is \(160^\circ\), so the triangle \(ABC\) is formed by two radii \(OB\) and \(OA\), and the chord \(BC\) and \(AC\). Wait, maybe the triangle \(OBC\) and \(OAC\) are isosceles, but actually, the key is that in triangle \(ABC\), \(OB = OC = OA\), so \(AB = AC\)? No, wait, the angle at \(O\) is \(160^\circ\), so the triangle \(OAB\) is isosceles with \(OB = OA\), but no, the angle at \(O\) is \(160^\circ\), so the base angles at \(B\) and \(A\) would be \(\frac{180 - 160}{2}=10^\circ\) each? Wait, no, that's not right. Wait, maybe the triangle \(ABC\) is an isosceles triangle with \(OB = OC = OA\), so \(BC = AC\)? No, let's start over.

Wait, the center is \(O\), so \(OB\) and \(OA\) are radii, so \(OB = OA\). The angle \(\angle AOB = 160^\circ\). Then, the triangle \(ABC\) has \(C\) on the circumference. Wait, actually, the angle at \(C\) (angle \(x\)) is an inscribed angle? No, wait, the triangle \(ABC\) is formed by two chords \(BC\) and \(AC\), and the central angle \(\angle AOB = 160^\circ\). Wait, maybe the triangle \(ABC\) is isosceles with \(AB = AC\), but no, let's use the property that the sum of angles in a triangle is \(180^\circ\), and the triangle \(OBC\) and \(OAC\) are isosceles. Wait, no, the correct approach is: in triangle \(ABC\), \(OB = OC = OA\), so \(OB = OC\) implies \(\angle OBC = \angle OCB\), and \(OA = OC\) implies \(\angle OAC = \angle OCA\). But maybe the triangle \(ABC\) is such that the angle at \(O\) is \(160^\circ\), so the remaining angle in the triangle \(OAB\) (wait, no, the triangle is \(ABC\) with \(O\) inside? Wait, no, the diagram shows \(O\) as the center, with points \(B\), \(O\), \(A\) in a line? No, \(B\), \(O\), \(A\) form a central angle of \(160^\circ\), and \(C\) is on the circumference. So triangle \(ABC\) with \(O\) as center, so \(OB = OC = OA\), so triangle \(OBC\) and \(OAC\) are isosceles. Wait, maybe the angle at \(C\) is \(x\), and we can find \(x\) by considering that the sum of angles in triangle \(ABC\) is \(180^\circ\), and the angle at \(O\) is \(160^\circ\), so the other two angles (at \(B\) and \(A\)) are equal? Wait, no, let's think of the triangle \(OBC\) and \(OAC\). Wait, \(OB = OC\), so \(\angle OBC = \angle OCB\), and \(OA = OC\), so \(\angle OAC = \angle OCA\). But the angle at \(O\) is \(160^\circ\), so the sum of angles in triangle \(ABC\) is \(180^\circ\). Wait, maybe the triangle \(ABC\) is isosceles with \(AB = AC\), so the base angles are equal. Wait, no, the key is that the central angle is \(160^\circ\), so the inscribed angle would be half, but that's for an arc. Wait, no, here we have a triangle with two radii and a chord. Wait, maybe the correct formula is that in a triangle with two sides as radii (so isosceles), the vertex angle is \(160^\circ\), so the base angles are \(\frac{180 - 160}{2}=10^\circ\) each? No, that can't be. Wait, no, the triangle here is \(ABC\), not \(OBC\). Wait, I think I made a mistake. Let's start over.

The center is \(O\), so \(OB = OC = OA\) (radii). So triangle \(OBC\) and \(OAC\) are isosceles. The angle at \(O\) ( \(\angle AOB\)) is \(160^\circ\). So in triangle \(OAB\), \(OB = OA\) (radii), so it's an isosceles triangle. The sum of angles in a triangle is \(180^\circ\), so the base angles at \(B\) and \(A\) are \(\frac{180 - 160}{2}=10^\circ\) each. But then, in triangle \(ABC\), we have angles at \(B\) and \(A\) as \(10^\circ\) each? No, that doesn't make sense. Wait, maybe the angle at \(C\) is \(x\), and we need to find \(x\) such that the sum of angles in triangle \(ABC\) is \(180^\circ\). Wait, no, the triangle \(ABC\) has angles at \(B\), \(A\), and \(C\) (which is \(x\)). But \(OB = OC\) and \(OA = OC\), so \(\angle OBC = \angle OCB\) and \(\angle OAC = \angle OCA\). Let's denote \(\angle OCB = y\) and \(\angle OCA = z\), so \(x = y + z\). Then, in triangle \(OBC\), \(\angle BOC = 180 - 2y\) (since it's isosceles), and in triangle \(OAC\), \(\angle AOC = 180 - 2z\). The sum of \(\angle BOC\) and \(\angle AOC\) is \(\angle AOB = 160^\circ\)? Wait, no, \(\angle AOB\) is \(160^\circ\), so \(\angle BOC + \angle AOC = 360 - 160 = 200^\circ\)? Wait, no, the total around point \(O\) is \(360^\circ\), so if \(\angle AOB = 160^\circ\), then the remaining angle is \(360 - 160 = 200^\circ\), which is \(\angle BOC + \angle AOC = 200^\circ\). But \(\angle BOC = 180 - 2y\) and \(\angle AOC = 180 - 2z\), so:

\((180 - 2y) + (180 - 2z) = 200\)

\(360 - 2(y + z) = 200\)

\(2(y + z) = 360 - 200 = 160\)

\(y + z = 80\)

But \(x = y + z\), so \(x = 80^\circ\). Wait, that makes sense. Let's check:

Total around \(O\): \(360^\circ\), so the other two angles ( \(\angle BOC\) and \(\angle AOC\)) sum to \(360 - 160 = 200^\circ\). Each of those angles is in an isosceles triangle ( \(OBC\) and \(OAC\) ) with two radii, so their base angles are equal. So for each triangle, the vertex angle (at \(O\)) is \(180 - 2 \times\) base angle. So sum of vertex angles is \(200^\circ\), so:

\((180 - 2y) + (180 - 2z) = 200\)

\(360 - 2(y + z) = 200\)

\(2(y + z) = 160\)

\(y + z = 80\)

Since \(x = y + z\), \(x = 80^\circ\).

Step2: Verify the calculation

We used the fact that the sum of angles around a point is \(360^\circ\), so the sum of \(\angle BOC\) and \(\angle AOC\) is \(360 - 160 = 200^\circ\). Then, using the isosceles triangle property (sum of angles in a triangle is \(180^\circ\)) for triangles \(OBC\) and \(OAC\), we derived that \(x = 80^\circ\).

Answer:

\(80\)