Question

what is the measure of \\(\overarc{dc}\\)?

options: 61, 119, 122, 360

Explanation:

Step1: Find angle at B

In triangle \(DBC\), sum of angles is \(180^\circ\). So \(\angle B = 180^\circ - 49^\circ - 70^\circ = 61^\circ\).

Step2: Relate inscribed angle to arc

The measure of an inscribed angle is half the measure of its intercepted arc. The arc \( \overarc{DC} \) is intercepted by inscribed angle \( \angle B \)? Wait, no—wait, actually, the inscribed angle theorem: the measure of an arc is twice the measure of the inscribed angle that intercepts it. Wait, let's correct. Wait, angle at \(B\) intercepts arc \(DC\)? Wait, no, angle at \(D\) is \(49^\circ\), intercepts arc \(BC\); angle at \(C\) is \(70^\circ\), intercepts arc \(BD\); angle at \(B\) intercepts arc \(DC\). Wait, the measure of arc \(DC\) is twice the measure of angle \(B\)? Wait, no, wait: in a circle, the measure of an inscribed angle is half the measure of its intercepted arc. So if angle at \(B\) is \(61^\circ\), then arc \(DC\) is \(2 \times 61^\circ = 122^\circ\)? Wait, no, wait—wait, maybe I mixed up. Wait, let's re - examine.

Wait, the triangle \(DBC\) is inscribed in the circle. So the sum of the measures of the arcs \(DB\), \(BC\), and \(DC\) is \(360^\circ\), but also, the measure of an inscribed angle is half the measure of its intercepted arc. Let's denote arc \(DB\) as \(x\), arc \(BC\) as \(y\), arc \(DC\) as \(z\). Then, angle at \(C\) (\(70^\circ\)) intercepts arc \(DB\), so \(70^\circ=\frac{1}{2}x\), so \(x = 140^\circ\). Angle at \(D\) (\(49^\circ\)) intercepts arc \(BC\), so \(49^\circ=\frac{1}{2}y\), so \(y = 98^\circ\). Then, since \(x + y+z=360^\circ\), \(140 + 98+z = 360\), \(z=360-(140 + 98)=122^\circ\). Yes, that makes sense. So arc \(DC\) is \(122^\circ\).

Wait, but earlier when we found angle at \(B\) is \(61^\circ\), and angle at \(B\) intercepts arc \(DC\), so arc \(DC = 2\times61 = 122^\circ\). That matches. So the measure of arc \(DC\) is \(122^\circ\).

Answer:

122