Angle in a Semicircle

Extended example

Problem. \(AB\) is a diameter of a circle with centre \(O\). A point \(C\) lies on the circle above the diameter. The tangent at \(A\) meets the extension of \(BC\) at \(T\). If \(\angle BAC=27^\circ\), find \(\angle ATB\).

Solution.

1. By the semicircle theorem, \(AB\) a diameter ⇒ \(\angle ACB=90^\circ\).
2. Triangle sum in \(\triangle ABC\): \(\angle ABC = 180^\circ-90^\circ-27^\circ = 63^\circ\).
3. Alternate Segment Theorem for chord \(AB\): the angle between the tangent at \(A\) and \(AB\) equals the angle in the alternate segment that subtends \(AB\). Since \(\angle ACB=90^\circ\), we get \(\angle TAB=90^\circ\).
4. Because \(T\) lies on the extension of \(BC\), the interior angle at \(B\) of \(\triangle ATB\) is \(\angle ABT = \angle ABC = 63^\circ\).
5. Now use the triangle angle sum in \(\triangle ATB\): \[ \angle ATB = 180^\circ - \angle TAB - \angle ABT = 180^\circ - 90^\circ - 63^\circ = 27^\circ. \]

Answer: \(\angle ATB = 27^\circ\).