Question

applying theorems what are the measures of arc jk and ∠kij? measure of arc jk = ° measure of ∠kij = °

Explanation:

Step1: Recall inscribed - angle theorem

The measure of an inscribed angle is half the measure of its intercepted arc. Let the measure of arc $\widehat{JK}$ be $x$. The inscribed angle $\angle KIJ$ intercepts arc $\widehat{JK}$, and another inscribed angle $\angle KJI$ intercepts arc $\widehat{KI}$ which has a measure of $116^{\circ}$, and $\angle JKI = 70^{\circ}$.

Step2: Use the angle - sum property of a triangle

In $\triangle JKI$, we know that the sum of the interior angles of a triangle is $180^{\circ}$. Let $\angle KIJ=\alpha$. Then $\alpha+70^{\circ}+\angle JKI = 180^{\circ}$. First, find $\angle JKI$. Since the measure of an inscribed angle $\angle JKI$ intercepts arc $\widehat{JI}$, and the measure of an inscribed angle $\angle KJI$ intercepts arc $\widehat{KI} = 116^{\circ}$, and $\angle JKI$ intercepts arc $\widehat{JI}$.
We know that the measure of an inscribed angle is half the measure of its intercepted arc.
The sum of the arcs of a circle is $360^{\circ}$. Let the measure of arc $\widehat{JK}=x$, arc $\widehat{KI} = 116^{\circ}$, and arc $\widehat{JI}=y$. So $x + y+116^{\circ}=360^{\circ}$, or $y=360^{\circ}-116^{\circ}-x$.
Also, using the angle - sum property of a triangle in $\triangle JKI$: $\angle KIJ+\angle JKI+\angle KJI = 180^{\circ}$. Since $\angle KJI$ is an inscribed angle intercepting arc $\widehat{KI}$, $\angle KJI=\frac{1}{2}\times116^{\circ}=58^{\circ}$.
Then $\angle KIJ=180^{\circ}-70^{\circ}-58^{\circ}=52^{\circ}$.
Since $\angle KIJ$ is an inscribed angle intercepting arc $\widehat{JK}$, and the measure of an inscribed angle is half the measure of its intercepted arc, if $\angle KIJ = 52^{\circ}$, then the measure of arc $\widehat{JK}=2\times\angle KIJ = 104^{\circ}$.

Answer:

Measure of $\widehat{JK}=104^{\circ}$
Measure of $\angle KIJ = 52^{\circ}$