Question

angle bcd is a circumscribed angle of circle a. angle bac measures 53°. what is the measure of angle bcd? 37° 106° 53° 74°

Explanation:

Step1: Analyze the quadrilateral \(AB - CD\)

Since \(BC\) and \(DC\) are tangent to the circle \(A\), \(AB\perp BC\) and \(AD\perp DC\) (tangent - radius property). So, \(\angle ABC=\angle ADC = 90^{\circ}\)

Step2: Use the sum of interior angles of a quadrilateral

The sum of interior angles of a quadrilateral is \(360^{\circ}\). In quadrilateral \(ABCD\), we know \(\angle ABC = 90^{\circ}\), \(\angle ADC=90^{\circ}\), \(\angle BAC = 53^{\circ}\), and let \(\angle BCD=x\)

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Another way: The central - angle \(\angle BAC\) and the circumscribed angle \(\angle BCD\) are related. The sum of the central angle \(\angle BAC\) and the circumscribed angle \(\angle BCD\) is \(180^{\circ}\) (theorem: the measure of a circumscribed angle and its corresponding central angle are supplementary)

$$ \angle BCD=180^{\circ}-\angle BAC $$

Step3: Calculate \(\angle BCD\)

Substitute \(\angle BAC = 106^{\circ}\) (Wait, no! Wait, correct formula: \(\angle BCD = 180^{\circ}- 2\angle BAC\) is wrong. Wait, correct: The measure of the circumscribed angle \(\angle BCD\) and the central angle \(\angle BAD\) (where \(\angle BAD = 2\angle BAC\) if \(AC\) is the angle - bisector). Wait, no, correct formula: \(\angle BCD=180^{\circ}-\angle BAD\). Since \(AB\perp BC\), \(AD\perp DC\), and \(AB = AD\) (radii of the same circle), \(\triangle ABC\cong\triangle ADC\) (by Hypotenuse - Leg congruence in right - triangles). So \(\angle BCA=\angle DCA\) and \(\angle BAC=\angle DAC\). \(\angle BAD = 2\angle BAC\). Using the property of the quadrilateral \(ABCD\) (\(AB\perp BC\), \(AD\perp DC\)), \(\angle BCD + \angle BAD=180^{\circ}\). Given \(\angle BAC = 53^{\circ}\), \(\angle BAD = 106^{\circ}\)

$$ \angle BCD=180^{\circ}-\angle BAD $$

$$ \angle BCD = 180^{\circ}-106^{\circ}=74^{\circ} $$

Answer:

\(74^{\circ}\)