Question

find the measures of the numbered angles in the kite.

Explanation:

Step1: Recall Kite Properties

A kite has two distinct pairs of adjacent sides equal. The diagonals of a kite are perpendicular, so \(\angle 1 = 90^\circ\). Also, one diagonal bisects the vertex angles, so \(\angle 3=\angle CAD\) and \(\triangle ABD\) is isosceles with \(AB = AD\), and \(\triangle ABC\cong\triangle ADC\) (SSS), so \(\angle 2=\angle CBD\)? Wait, no, let's correct. The diagonal \(AC\) bisects \(\angle BAD\), so \(\angle 3 = 55^\circ\)? Wait, no, the given angle at \(A\) is \(55^\circ\) for \(\angle BAD\)? Wait, the diagram: \(AB = AD\) (marked), \(BC = CD\) (marked). So diagonal \(AC\) bisects \(\angle BAD\) and diagonal \(BD\) is bisected by \(AC\) at right angles? Wait, no, in a kite, one diagonal is perpendicular to the other. So \(\angle 1 = 90^\circ\) (since diagonals of kite are perpendicular). Then, \(\angle 3\): since \(AB = AD\), \(\triangle ABD\) is isosceles, and \(AC\) bisects \(\angle BAD\), so \(\angle 3 = 55^\circ\)? Wait, no, the angle at \(A\) between \(AD\) and the diagonal is \(55^\circ\), so \(\angle 3\) is equal to that? Wait, maybe the given angle is \(\angle CAD = 55^\circ\), so \(\angle 3 = 55^\circ\)? Wait, no, let's re-express.

Wait, the problem is to find \(\angle 1\), \(\angle 2\), \(\angle 3\). Let's start with \(\angle 1\): in a kite, the diagonals are perpendicular, so \(\angle 1 = 90^\circ\). Then, \(\angle 3\): since \(AB = AD\), \(\triangle ABD\) is isosceles, and \(AC\) is the angle bisector, so \(\angle 3 = 55^\circ\) (because the given angle at \(A\) is \(55^\circ\) for \(\angle CAD\), so \(\angle 3 = 55^\circ\)). Then, for \(\angle 2\): in \(\triangle ADC\), we have \(\angle 1 = 90^\circ\), \(\angle CAD = 55^\circ\), so \(\angle 2 = 180^\circ - 90^\circ - 55^\circ = 35^\circ\). Wait, let's verify:

1. \(\angle 1\): Diagonals of kite are perpendicular, so \(\angle 1 = 90^\circ\).
2. \(\angle 3\): Since \(AB = AD\) and \(AC\) bisects \(\angle BAD\), \(\angle 3 = 55^\circ\) (given angle at \(A\) is \(55^\circ\) for \(\angle CAD\), so \(\angle 3\) is equal to that? Wait, maybe the angle at \(A\) is \(\angle BAD = 110^\circ\), and \(AC\) bisects it, so \(\angle 3 = 55^\circ\). Wait, maybe the given angle is \(\angle CAD = 55^\circ\), so \(\angle 3 = 55^\circ\). Then, in \(\triangle ADC\), right-angled at \(C\) (wait, no, \(\angle 1 = 90^\circ\), so \(\triangle ADC\) has \(\angle 1 = 90^\circ\), \(\angle CAD = 55^\circ\), so \(\angle 2 = 180 - 90 - 55 = 35^\circ\). And \(\angle 1 = 90^\circ\), \(\angle 3 = 55^\circ\), \(\angle 2 = 35^\circ\).

Wait, let's structure:

Step1: Find \(\angle 1\)

In a kite, the diagonals are perpendicular. So \(\angle 1 = 90^\circ\).

Step2: Find \(\angle 3\)

Since \(AB = AD\) (marked), \(\triangle ABD\) is isosceles. The diagonal \(AC\) bisects \(\angle BAD\), so \(\angle 3 = 55^\circ\) (given angle at \(A\) is \(55^\circ\) for \(\angle CAD\), so \(\angle 3\) equals that).

Step3: Find \(\angle 2\)

In \(\triangle ADC\), we know \(\angle 1 = 90^\circ\) and \(\angle CAD = 55^\circ\). Using the angle sum property of a triangle (\(180^\circ\)):
\(\angle 2 = 180^\circ - \angle 1 - \angle CAD = 180^\circ - 90^\circ - 55^\circ = 35^\circ\).

Answer:

\(\angle 1 = 90^\circ\), \(\angle 2 = 35^\circ\), \(\angle 3 = 55^\circ\)