Question

ABCD is a quadrilateral in which \(\angle D=60^\circ\) and \(\angle C=100^\circ\). If the internal bisectors of \(\angle A\) and \(\angle B\) meet at \(P\), then the measure of \(\angle APB\) is:

• A. \(80^\circ\)
• B. \(90^\circ\)
• C. \(100^\circ\)
• D. \(110^\circ\)

Hint:

For adjacent angles in a polygon sharing a side, the angle between their internal bisectors equals \(180^\circ-\dfrac{\text{sum of those angles}}{2}\). In a quadrilateral this can be written as \(\dfrac{\text{sum of the other two angles}}{2}\).

Answer 1

The Correct Option is A

Step 1: Use angle sum of a quadrilateral.
\(\angle A+\angle B+\angle C+\angle D=360^\circ\). Given \(\angle C=100^\circ\) and \(\angle D=60^\circ\), hence \[ \angle A+\angle B=360^\circ-(100^\circ+60^\circ)=200^\circ. \] Step 2: Relate \(\angle APB\) to \(\angle A\) and \(\angle B\).
The angle between the internal bisectors of two adjacent angles \(\angle A\) and \(\angle B\) at vertices sharing side \(AB\) equals \[ \angle APB=180^\circ-\frac{\angle A+\angle B}{2}. \] (Reason: from the bisector at \(A\) to side \(AB\) is \(\frac{\angle A}{2}\); then turning along the straight line from \(AB\) to \(BA\) adds \(180^\circ\); from \(BA\) to the bisector at \(B\) adds \(\frac{\angle B}{2}\). The smaller angle is \(360^\circ-\left(180^\circ+\frac{\angle A+\angle B}{2}\right)\).) Step 3: Compute \(\angle APB\).
\[ \angle APB=180^\circ-\frac{200^\circ}{2}=180^\circ-100^\circ=80^\circ. \] (Equivalently, \(\angle APB=\frac{\angle C+\angle D}{2}=\frac{100^\circ+60^\circ}{2}=80^\circ\).) \[ \boxed{80^\circ} \]