Question

lm, ∠k = 56° and ∠m = 44°. what is the measure of ∠kom? select one answer a 40° b 80° c 100° d 130°

Explanation:

Step1: Analyze right triangles

In right triangle \( \triangle KHM \), \( \angle K = 56^\circ \), \( \angle H = 90^\circ \), so \( \angle HKM = 90^\circ - 56^\circ = 34^\circ \)? Wait, no, wait. Wait, \( \angle K \) is \( 56^\circ \), \( \angle M \) is \( 44^\circ \), so in triangle \( KLM \), \( \angle L = 180 - 56 - 44 = 80^\circ \). Then, \( KH \perp LM \), \( MG \perp LM \)? Wait, \( KH \) and \( MG \) are altitudes? Wait, \( \angle OHK = 90^\circ \), \( \angle OGM = 90^\circ \). Then, quadrilateral \( HOGM \) has two right angles. Wait, maybe better to look at the angles at \( O \).

Wait, \( \angle K = 56^\circ \), \( \angle M = 44^\circ \), so the angles at \( K \) and \( M \) in triangle \( KLM \). Then, \( KH \) and \( MG \) are perpendicular to \( LM \), so \( \angle KHO = \angle MGO = 90^\circ \). Then, in triangle \( KOH \) and \( MOG \), but maybe the sum of angles around \( O \). Wait, alternatively, the measure of \( \angle KOM \) can be found by considering the sum of angles. Wait, \( \angle K = 56^\circ \), so the angle at \( K \) in the right triangle \( KHL \): \( \angle HKL = 56^\circ \), so \( \angle HKO = 56^\circ \)? No, wait, \( \angle K = 56^\circ \) is \( \angle LKM = 56^\circ \), and \( \angle LMK = 44^\circ \). Then, \( KH \) is perpendicular to \( LM \), so \( \angle KHL = 90^\circ \), so in triangle \( KHL \), \( \angle HKL = 56^\circ \), \( \angle KHL = 90^\circ \), so \( \angle KLH = 34^\circ \)? No, that contradicts earlier. Wait, no, \( \angle K = 56^\circ \) is \( \angle LKM = 56^\circ \), \( \angle M = 44^\circ \) is \( \angle LMK = 44^\circ \), so \( \angle L = 180 - 56 - 44 = 80^\circ \). Then, \( KH \) and \( MG \) are altitudes, so \( \angle KHL = \angle MGL = 90^\circ \). Then, quadrilateral \( HOGM \) has \( \angle H = 90^\circ \), \( \angle G = 90^\circ \), \( \angle L = 80^\circ \), so \( \angle HOG = 180 - 80 = 100^\circ \)? Wait, no, \( \angle HOG \) is equal to \( \angle KOM \) (vertical angles). Wait, maybe:

In triangle \( KLM \), \( \angle L = 80^\circ \). \( KH \perp LM \), \( MG \perp LM \), so \( KH \parallel MG \) (both perpendicular to \( LM \)). Then, the angle between \( KO \) and \( MO \) is \( 180 - (90 - 56) - (90 - 44) \)? Wait, \( \angle HKO = 90 - 56 = 34^\circ \), \( \angle GMO = 90 - 44 = 46^\circ \)? No, that doesn't add up. Wait, maybe the correct approach is:

The sum of angles in a quadrilateral is \( 360^\circ \). In quadrilateral \( HOGM \), \( \angle H = 90^\circ \), \( \angle G = 90^\circ \), \( \angle L = 80^\circ \) (since \( \angle L \) is equal to \( \angle HOG \) because \( KH \parallel MG \) and \( LM \) is a transversal? No, wait, \( \angle HOG \) and \( \angle L \) are same-side interior angles? Wait, no, \( KH \) and \( MG \) are both perpendicular to \( LM \), so they are parallel. Then, \( LM \) is the transversal, so \( \angle L \) and \( \angle HOG \) are same-side interior angles, so they should be supplementary. Wait, \( \angle L = 80^\circ \), so \( \angle HOG = 180 - 80 = 100^\circ \). But \( \angle HOG \) is equal to \( \angle KOM \) (vertical angles), so \( \angle KOM = 100^\circ \)? But the options are 40, 80, 100, 130. Wait, 100 is option C. But wait, maybe I made a mistake.

Wait, let's re-examine. \( \angle K = 56^\circ \), \( \angle M = 44^\circ \), so in triangle \( KLM \), \( \angle L = 80^\circ \). \( KH \perp LM \), so \( \angle KHL = 90^\circ \), so in triangle \( KHL \), \( \angle HKL = 56^\circ \), so \( \angle KLH = 34^\circ \)? No, that's not right. Wait, \( \angle K \) is \( 56^\circ \), which is \( \angle LKM = 56^\circ \), so in triangle \( KHL \), \( \angle LKM = 56^\circ \), \( \angle KHL = 90^\circ \), so \( \angle KLH = 180 - 56 - 90 = 34^\circ \). Similarly, in triangle \( MGL \), \( \angle LMK = 44^\circ \), \( \angle MGL = 90^\circ \), so \( \angle MLG = 180 - 44 - 90 = 46^\circ \). Then, \( \angle KLH + \angle MLG = 34 + 46 = 80^\circ \), which is \( \angle L \), so that checks out. Then, the angle at \( O \): in quadrilateral \( HOGM \), \( \angle H = 90^\circ \), \( \angle G = 90^\circ \), \( \angle L = 80^\circ \), so \( \angle HOG = 360 - 90 - 90 - 80 = 100^\circ \). So \( \angle KOM = 100^\circ \), which is option C.

Wait, but let's check again. The sum of angles around point \( O \): \( \angle KOM + \angle KOH + \angle MOG = 360 \), but no, \( \angle KOH \) and \( \angle MOG \) are adjacent to \( \angle KOM \). Alternatively, the measure of \( \angle KOM \) is equal to \( 180^\circ - \angle L \)? No, \( \angle L = 80^\circ \), so \( 180 - 80 = 100 \), which matches. So the answer should be 100 degrees, option C.

Step1: Calculate \( \angle L \) in \( \triangle KLM \)

In \( \triangle KLM \), the sum of angles is \( 180^\circ \). Given \( \angle K = 56^\circ \) and \( \angle M = 44^\circ \), we calculate:
\( \angle L = 180^\circ - \angle K - \angle M = 180^\circ - 56^\circ - 44^\circ = 80^\circ \).

Step2: Analyze quadrilateral \( HOGM \)

\( KH \perp LM \) and \( MG \perp LM \), so \( \angle OHK = 90^\circ \) and \( \angle OGM = 90^\circ \). In quadrilateral \( HOGM \), the sum of interior angles is \( 360^\circ \). Let \( \angle HOG = \angle KOM \) (vertical angles). Using the angle sum:
\( \angle HOG + \angle OHK + \angle OGM + \angle L = 360^\circ \).
Substituting known angles:
\( \angle HOG + 90^\circ + 90^\circ + 80^\circ = 360^\circ \).
Simplify:
\( \angle HOG = 360^\circ - 90^\circ - 90^\circ - 80^\circ = 100^\circ \).

Since \( \angle HOG = \angle KOM \) (vertical angles), \( \angle KOM = 100^\circ \).

Answer:

C. \( 100^\circ \)