Question

in the circle below, suppose m\\(\overarc{kli}\\)=248° and m\\(\angle lkj\\)=131°. find the following. (a) m\\(\angle kli\\) = \\(\square\\)° (b) m\\(\angle lij\\) = \\(\square\\)°

Explanation:

Step1: Find the measure of the minor arc KI

The total circumference of a circle is \(360^\circ\). Given \(m\overarc{KLI} = 248^\circ\), the measure of the minor arc \(KI\) is \(360^\circ - 248^\circ = 112^\circ\).

Step2: Find \(m\angle KLI\) (a)

In a cyclic quadrilateral, the measure of an inscribed angle is half the measure of its intercepted arc. For \(\angle KLI\), it intercepts arc \(KJ I\)? Wait, no. Wait, actually, in a cyclic quadrilateral, opposite angles are supplementary? Wait, no, let's recall. Wait, the inscribed angle theorem: the measure of an inscribed angle is half the measure of its intercepted arc. Wait, \(\angle KLI\) intercepts arc \(KJ I\)? Wait, no, the arc \(KI\) is \(112^\circ\), and \(\angle KLI\) is an inscribed angle intercepting arc \(KJ I\)? Wait, maybe I made a mistake. Wait, the problem says \(m\overarc{KLI} = 248^\circ\), so the arc \(KI\) is \(360 - 248 = 112^\circ\). Now, \(\angle KLI\): wait, maybe \(\angle KLI\) is an inscribed angle intercepting arc \(KJ I\)? No, wait, let's look at the cyclic quadrilateral \(LKJI\). Wait, in a cyclic quadrilateral, the angle subtended by an arc at the circumference is half the arc. Wait, maybe \(\angle KLI\) intercepts arc \(KJ I\), but arc \(KJ I\) is \(248^\circ\)? No, that can't be. Wait, no, the inscribed angle theorem: the measure of an inscribed angle is half the measure of its intercepted arc. So if \(\angle KLI\) intercepts arc \(KI\), then \(m\angle KLI = \frac{1}{2}m\overarc{KI}\). Wait, but arc \(KI\) is \(112^\circ\), so \(\frac{1}{2} \times 112^\circ = 56^\circ\)? Wait, no, maybe I messed up. Wait, the problem says \(m\angle LKJ = 131^\circ\). Let's check the cyclic quadrilateral. In a cyclic quadrilateral, opposite angles are supplementary. Wait, \(LKJI\) is a cyclic quadrilateral, so \(\angle LKJ + \angle LIJ = 180^\circ\), and \(\angle KLI + \angle KJI = 180^\circ\)? Wait, no, maybe not. Wait, let's re-examine.

Wait, the arc \(KLI\) is \(248^\circ\), so the arc \(KI\) (the minor arc) is \(360 - 248 = 112^\circ\). Now, \(\angle KLI\) is an inscribed angle that intercepts arc \(KI\)? Wait, no, \(\angle KLI\) is at point \(L\), with sides \(LK\) and \(LI\), so it intercepts arc \(KI\). So by the inscribed angle theorem, \(m\angle KLI = \frac{1}{2}m\overarc{KI} = \frac{1}{2} \times 112^\circ = 56^\circ\). Wait, but let's check the other angle. \(\angle LKJ = 131^\circ\), which is an inscribed angle intercepting arc \(LJI\). Arc \(LJI\) is \(248^\circ\), so \(\frac{1}{2} \times 248^\circ = 124^\circ\), but that's not \(131^\circ\). So maybe my initial approach is wrong.

Wait, maybe the quadrilateral is cyclic, so opposite angles are supplementary. So \(\angle LKJ + \angle LIJ = 180^\circ\), so \(131^\circ + \angle LIJ = 180^\circ\), so \(\angle LIJ = 49^\circ\). Wait, but that's for part (b). Wait, let's do part (a) first.

Wait, the arc \(KLI\) is \(248^\circ\), so the arc \(KI\) is \(360 - 248 = 112^\circ\). Now, \(\angle KLI\) is an inscribed angle intercepting arc \(KI\), so \(m\angle KLI = \frac{1}{2} \times 112^\circ = 56^\circ\). Wait, but let's check with the cyclic quadrilateral. In cyclic quadrilateral \(LKJI\), \(\angle KLI + \angle KJI = 180^\circ\), and \(\angle LKJ + \angle LIJ = 180^\circ\). Wait, maybe. Let's confirm.

Alternatively, maybe the angle \(\angle KLI\) is an inscribed angle intercepting arc \(KJ I\), but arc \(KJ I\) is \(248^\circ\), so \(\frac{1}{2} \times 248^\circ = 124^\circ\), but that contradicts. Wait, no, the inscribed angle intercepts the arc that is opposite to the angle. Wait, maybe I need to draw the circle. Points \(L\), \(K\), \(J\), \(I\) are on the circle. So \(L\) to \(K\) to \(J\) to \(I\) to \(L\) is the quadrilateral. Arc \(KLI\) is from \(K\) to \(L\) to \(I\), which is \(248^\circ\), so the arc from \(K\) to \(I\) through \(J\) is \(360 - 248 = 112^\circ\). Now, \(\angle KLI\) is at \(L\), between \(K\) and \(I\), so it intercepts arc \(KI\) (the minor arc, \(112^\circ\)), so the angle is half of that, \(56^\circ\). Then for part (b), \(\angle LIJ\): let's see, \(\angle LIJ\) is at \(I\), between \(L\) and \(J\). It intercepts arc \(L K\)? Wait, no. Wait, \(\angle LIJ\) intercepts arc \(L K\)? Wait, no, the arc \(L K J\)? Wait, maybe using the cyclic quadrilateral: in cyclic quadrilateral \(LKJI\), \(\angle LKJ + \angle LIJ = 180^\circ\) (opposite angles are supplementary). Given \(m\angle LKJ = 131^\circ\), so \(m\angle LIJ = 180^\circ - 131^\circ = 49^\circ\).

Wait, let's verify part (a) again. If \(\angle KLI\) is \(56^\circ\), and \(\angle KJI\) would be \(180 - 56 = 124^\circ\), but does that make sense? Let's check the arc. Arc \(KLI\) is \(248^\circ\), so the inscribed angle over arc \(KLI\) would be \(124^\circ\) (half of \(248\)), which would be \(\angle KJI\), since \(\angle KJI\) intercepts arc \(KLI\). So \(\angle KJI = \frac{1}{2} \times 248^\circ = 124^\circ\), then \(\angle KLI = 180 - 124 = 56^\circ\) (since they are opposite angles in cyclic quadrilateral). Yes, that works. So part (a) is \(56^\circ\), part (b) is \(49^\circ\).

Answer:

(a) \(m\angle KLI = \boxed{56}\) degrees
(b) \(m\angle LIJ = \boxed{49}\) degrees