Question

in the circle below, \\( \overline{ik} \\) is a diameter. suppose \\( m\overarc{jk} = 118^\circ \\) and \\( m\angle kjl = 24^\circ \\). find the following.
(a) \\( m\angle ijl = \square^\circ \\)
(b) \\( m\angle ikj = \square^\circ \\)

Explanation:

Step1: Find \( m\angle IJL \)

Since \( \overline{IK} \) is a diameter, the measure of arc \( IK \) is \( 180^\circ \). We know \( m\overarc{JK} = 118^\circ \), so the measure of arc \( IJ \) is \( 180^\circ - 118^\circ = 62^\circ \). The inscribed angle theorem states that an inscribed angle is half the measure of its intercepted arc. For \( \angle IJL \), it intercepts arc \( IL \)? Wait, no. Wait, \( \angle KJL = 24^\circ \), and we can find \( \angle IJL \) by first finding the inscribed angle over arc \( IJ \). Wait, actually, \( \angle IJK \) is an inscribed angle over arc \( IJ \), so \( m\angle IJK = \frac{1}{2}m\overarc{IJ} = \frac{1}{2} \times 62^\circ = 31^\circ \)? Wait, no, maybe another approach. Wait, \( \overline{IK} \) is diameter, so \( \angle IJK \) is an inscribed angle? Wait, no, \( \triangle IJK \) is a triangle with \( \overline{IK} \) as diameter, so \( \angle IJK = 90^\circ \)? Wait, no, that's Thales' theorem: an angle inscribed in a semicircle is a right angle. So \( \angle IJK = 90^\circ \). Then, since \( \angle KJL = 24^\circ \), then \( \angle IJL = \angle IJK - \angle KJL = 90^\circ - 24^\circ = 66^\circ \)? Wait, no, maybe I messed up. Wait, let's re-examine.

Wait, the arc \( JK \) is \( 118^\circ \), so the inscribed angle over arc \( JK \) would be \( \frac{1}{2} \times 118^\circ = 59^\circ \). Wait, \( \angle IKJ \) is an inscribed angle over arc \( IJ \). Wait, arc \( IJ \) is \( 180^\circ - 118^\circ = 62^\circ \), so \( \angle IKJ = \frac{1}{2} \times 62^\circ = 31^\circ \). Then, for part (a), \( \angle IJL \): since \( \angle KJL = 24^\circ \), and \( \angle IJK = 90^\circ \) (Thales' theorem, because \( \overline{IK} \) is diameter, so \( \angle IJK \) is right angle), then \( \angle IJL = \angle IJK - \angle KJL = 90^\circ - 24^\circ = 66^\circ \)? Wait, no, maybe Thales' theorem: \( \angle IJK = 90^\circ \), so \( \angle IJL + \angle KJL = 90^\circ \), so \( \angle IJL = 90^\circ - 24^\circ = 66^\circ \).

Wait, let's correct. Thales' theorem: if you have a triangle inscribed in a circle where one side is the diameter, then the angle opposite that side is a right angle. So \( \triangle IJK \) has \( \overline{IK} \) as diameter, so \( \angle IJK = 90^\circ \). Therefore, \( \angle IJL + \angle KJL = 90^\circ \). Given \( \angle KJL = 24^\circ \), so \( \angle IJL = 90^\circ - 24^\circ = 66^\circ \).

For part (b), \( \angle IKJ \): the arc \( IJ \) is \( 180^\circ - 118^\circ = 62^\circ \), so the inscribed angle \( \angle IKJ \) intercepts arc \( IJ \), so \( m\angle IKJ = \frac{1}{2} \times 62^\circ = 31^\circ \).

Wait, let's verify:

Arc \( JK = 118^\circ \), so inscribed angle \( \angle IKJ \) intercepts arc \( IJ \), which is \( 180 - 118 = 62 \), so \( \angle IKJ = 31^\circ \). Then, in \( \triangle IJK \), angles sum to \( 180^\circ \), so \( \angle IJK = 90^\circ \), \( \angle IKJ = 31^\circ \), so \( \angle JIK = 180 - 90 - 31 = 59^\circ \), which is half of arc \( JK \) (118/2=59), so that checks out. Then, \( \angle KJL = 24^\circ \), which is an inscribed angle over arc \( KL \). So arc \( KL \) is \( 2 \times 24^\circ = 48^\circ \). Then arc \( IL \) is arc \( IK - arc KL = 180 - 48 = 132^\circ \)? No, maybe not. Wait, back to part (a): \( \angle IJL \) intercepts arc \( IL \)? Wait, no, \( \angle IJL \) is formed by chords \( IJ \) and \( JL \), so it intercepts arc \( IL \). Wait, but maybe my initial approach with Thales' theorem was wrong. Wait, let's start over.

1. \( \overline{IK} \) is diameter, so \( m\overarc{IK} = 180^\circ \).
2. \( m\overarc{JK} = 118^\circ \), so \( m\overarc{IJ} = 180^\circ - 118^\circ = 62^\circ \).
3. The inscribed angle over arc \( IJ \) is \( \angle IKJ \), so \( m\angle IKJ = \frac{1}{2}m\overarc{IJ} = \frac{1}{2} \times 62^\circ = 31^\circ \). (This is part (b))
4. For part (a), \( \angle KJL = 24^\circ \), which is an inscribed angle over arc \( KL \), so \( m\overarc{KL} = 2 \times 24^\circ = 48^\circ \).
5. Then, arc \( IL \) is \( m\overarc{IK} - m\overarc{KL} = 180^\circ - 48^\circ = 132^\circ \)? No, that doesn't make sense. Wait, no, \( \angle IJL \) is an inscribed angle over arc \( IL \), so \( m\angle IJL = \frac{1}{2}m\overarc{IL} \). But we need to find \( m\overarc{IL} \). Wait, arc \( IJ \) is 62°, arc \( JK \) is 118°, arc \( KL \) is 48° (from \( \angle KJL = 24^\circ \)), so arc \( IL \) is arc \( IJ + arc JK + arc KL \)? No, the circle is 360°, but \( \overline{IK} \) is diameter, so the semicircle is 180°. Wait, the points are K, L, I on the circle? Wait, the diagram shows K, L, I on the circle with \( \overline{IK} \) as diameter, and J above. So the semicircle from K to I is 180°, with J somewhere on the upper half. So arc JK is 118°, so arc IJ is 62°, as before. Then, angle \( \angle KJL = 24^\circ \) is an inscribed angle, so arc KL is 48°, so arc IL is arc IK - arc KL = 180 - 48 = 132°, so \( \angle IJL = \frac{1}{2} \times 132^\circ = 66^\circ \). Yes, that matches the Thales' theorem approach. So \( \angle IJL = 66^\circ \), \( \angle IKJ = 31^\circ \).

Step2: Confirm part (b)

As calculated, \( m\overarc{IJ} = 62^\circ \), so the inscribed angle \( \angle IKJ \) (which intercepts arc \( IJ \)) is \( \frac{1}{2} \times 62^\circ = 31^\circ \).

Answers:

(a) \( \boldsymbol{66} \)
(b) \( \boldsymbol{31} \)

Answer:

s:
(a) \( \boldsymbol{66} \)
(b) \( \boldsymbol{31} \)