Question

what are the measures of \\(\overarc{jk}\\) and \\(\angle kij\\)? measure of \\(\overarc{jk} = \square^\circ\\) measure of \\(\angle kij = \square^\circ\\) (with a circle diagram having points k, i, j, angle at k is 70°, arc from k to i is 116°)

Explanation:

Step1: Find measure of arc JK

The total degrees in a circle is \( 360^\circ \). We know one arc is \( 116^\circ \) and the inscribed angle \( \angle KJI \) is related, but first, recall that the measure of an inscribed angle is half the measure of its intercepted arc. Wait, actually, to find arc \( JK \), we can use the fact that the sum of arcs in a circle is \( 360^\circ \), but also, the inscribed angle \( \angle KIJ \) or wait, no, let's look at the given angle. Wait, the triangle at K has an angle of \( 70^\circ \), but maybe better to use the fact that the measure of an inscribed angle is half the intercepted arc. Wait, no, let's first find arc \( JK \). Wait, the arc \( KI \) is \( 116^\circ \)? Wait, no, the diagram shows arc \( KI \) as \( 116^\circ \)? Wait, no, the label is \( 116^\circ \) on the arc from K to I? Wait, maybe the circle has three arcs: arc JK, arc KI, and arc IJ? Wait, no, the points are J, K, I on the circle. So the arcs are JK, KI, and IJ. The sum of arcs in a circle is \( 360^\circ \). Wait, but we also know that the inscribed angle \( \angle KJI \) or \( \angle KIJ \)? Wait, the angle at K is \( 70^\circ \), which is an inscribed angle? Wait, no, the angle at K is \( 70^\circ \), which is an inscribed angle intercepting arc IJ. Wait, the measure of an inscribed angle is half the measure of its intercepted arc. So if \( \angle K = 70^\circ \), then arc IJ is \( 2 \times 70^\circ = 140^\circ \). Then, we know arc KI is \( 116^\circ \) (from the diagram), so arc JK can be found by \( 360^\circ - arc IJ - arc KI \). Wait, \( 360 - 140 - 116 = 104^\circ \)? Wait, no, that can't be. Wait, maybe I got the arcs wrong. Wait, let's re-examine. The points are J, K, I on the circle. So the arcs are JK, KI, and IJ. The angle at K is \( 70^\circ \), which is an inscribed angle intercepting arc IJ. So arc IJ = \( 2 \times 70^\circ = 140^\circ \). The arc KI is given as \( 116^\circ \)? Wait, the diagram shows \( 116^\circ \) on the arc from K to I. Then arc JK = \( 360^\circ - arc IJ - arc KI = 360 - 140 - 116 = 104^\circ \)? Wait, no, that seems off. Wait, maybe the \( 116^\circ \) is arc IJ? No, the label is near K and I. Wait, maybe the angle at K is \( 70^\circ \), which is an inscribed angle, so arc IJ is \( 140^\circ \), and arc KI is \( 116^\circ \), so arc JK is \( 360 - 140 - 116 = 104 \)? Wait, but let's check the second part, the measure of \( \angle KIJ \). The measure of an inscribed angle is half the measure of its intercepted arc. So \( \angle KIJ \) intercepts arc JK. So if arc JK is \( 104^\circ \), then \( \angle KIJ = \frac{1}{2} \times 104^\circ = 52^\circ \). Wait, but let's verify. Alternatively, the sum of angles in a triangle is \( 180^\circ \). If \( \angle K = 70^\circ \), and \( \angle KIJ = x \), then \( \angle KJI = 180 - 70 - x = 110 - x \). But \( \angle KJI \) is an inscribed angle intercepting arc KI, which is \( 116^\circ \), so \( \angle KJI = \frac{1}{2} \times 116^\circ = 58^\circ \). Then, \( 180 - 70 - 58 = 52^\circ \), so \( \angle KIJ = 52^\circ \). Then, arc JK is intercepted by \( \angle KIJ \), so arc JK = \( 2 \times 52^\circ = 104^\circ \). Yes, that makes sense. So:

Step1: Find arc JK

First, find \( \angle KJI \), which is an inscribed angle intercepting arc KI (116°). So \( \angle KJI = \frac{1}{2} \times 116^\circ = 58^\circ \).

Step2: Find \( \angle KIJ \)

In triangle KIJ, sum of angles is 180°. So \( \angle KIJ = 180^\circ - 70^\circ - 58^\circ = 52^\circ \).

Step3: Find arc JK

\( \angle KIJ \) is an inscribed angle intercepting arc JK, so arc JK = \( 2 \times \angle KIJ = 2 \times 52^\circ = 104^\circ \).

Answer:

Measure of \( \widehat{JK} = \boxed{104}^\circ \)
Measure of \( \angle KIJ = \boxed{52}^\circ \)