Question

solve for m<nlm.

circle with center, points l (top), n (bottom), m (right). ln is diameter. lm arc: 13x - 10, nm arc: 7x - 10. multiple choice options: 10°, 30°, 45°, 60°

Explanation:

Step1: Identify the triangle type

Since \( LN \) is a diameter, \( \angle LMN = 90^\circ \) (Thales' theorem), and \( \triangle LMN \) is a right triangle. Also, \( LM \) and \( MN \) are chords, and the arcs \( LM \) and \( MN \) correspond to angles. Wait, actually, in a circle, if \( LN \) is diameter, then \( \angle LMN = 90^\circ \), and the arcs: arc \( LM = 13x - 10 \), arc \( MN = 7x - 10 \). The sum of arcs \( LM \) and \( MN \) should be \( 180^\circ \) (since \( LN \) is diameter, semicircle is \( 180^\circ \)). So:

$$ (13x - 10) + (7x - 10) = 180 $$

Step2: Solve for \( x \)

Combine like terms:

$$ 20x - 20 = 180 $$

Add 20 to both sides:

$$ 20x = 200 $$

Divide by 20:

$$ x = 10 $$

Step3: Find arc \( MN \)

Arc \( MN = 7x - 10 = 7(10) - 10 = 60^\circ \). The inscribed angle \( \angle NLM \) subtends arc \( MN \), so \( m\angle NLM=\frac{1}{2}\times \text{arc } MN \)? Wait, no: inscribed angle subtended by arc \( MN \) is \( \angle LMN \)? Wait, no. Wait, \( \angle NLM \) is an inscribed angle subtended by arc \( MN \). Wait, in \( \triangle LMN \), right-angled at \( M \), \( \angle NLM \) and \( \angle LNM \) are complementary. Wait, arc \( LM \) is \( 13x - 10 = 13(10)-10 = 120^\circ \), arc \( MN = 60^\circ \). Then \( \angle NLM \) is an inscribed angle subtended by arc \( MN \), so \( m\angle NLM=\frac{1}{2}\times \text{arc } MN \)? No, wait: inscribed angle subtended by arc \( MN \) is \( \angle LMN \)? No, I think I mixed up. Wait, \( \angle NLM \) is at \( L \), so it subtends arc \( MN \). The measure of an inscribed angle is half the measure of its subtended arc. So arc \( MN = 60^\circ \), so \( m\angle NLM=\frac{1}{2}\times 60^\circ = 30^\circ \)? Wait, no: wait, \( \angle NLM \) subtends arc \( MN \), so yes, \( m\angle NLM=\frac{1}{2}\times \text{arc } MN \). Wait, arc \( MN = 7x - 10 = 60^\circ \), so \( \angle NLM = \frac{1}{2} \times 60^\circ = 30^\circ \). Let's check: arc \( LM = 13x - 10 = 120^\circ \), so \( \angle LNM \) subtends arc \( LM \), so \( \angle LNM = \frac{1}{2} \times 120^\circ = 60^\circ \). Then in right triangle \( LMN \), \( \angle NLM + \angle LNM = 90^\circ \), so \( 30^\circ + 60^\circ = 90^\circ \), which works. So \( m\angle NLM = 30^\circ \).

Answer:

\( 30^\circ \) (corresponding to the option with \( 30^\circ \))