Question

in the diagram below, point d is the center of the circle, and the measure of \\(\overarc{bc}\\) is \\(50^{\circ}\\). what is the measure of \\(\angle acd\\)? \\(\bigcirc\\ 50^{\circ}\\) \\(\bigcirc\\ 65^{\circ}\\) \\(\bigcirc\\ 25^{\circ}\\) \\(\bigcirc\\ 40^{\circ}\\)

Explanation:

Step1: Identify central angle

Since \( D \) is the center, \( \angle BDC \) is a central angle equal to the measure of arc \( BC \), so \( \angle BDC = 50^\circ \).

Step2: Recognize isosceles triangle

\( AD = CD \) (radii of the circle), so \( \triangle ACD \) is isosceles with \( \angle A = \angle ACD \).

Step3: Find \( \angle ADC \)

\( \angle ADC \) and \( \angle BDC \) are supplementary (linear pair), so \( \angle ADC = 180^\circ - 50^\circ = 130^\circ \).

Step4: Calculate \( \angle ACD \)

In \( \triangle ACD \), the sum of angles is \( 180^\circ \). Let \( \angle ACD = x \), then \( x + x + 130^\circ = 180^\circ \). Solving: \( 2x = 50^\circ \), so \( x = 65^\circ \).

Answer:

65° (corresponding to the option with 65°)