Question

determine if triangle jkl and triangle mno are or are not similar, and, if they are, state how you know. (note that figures are not necessarily drawn to scale.)

Explanation:

Step1: Find angle in triangle MNO

In triangle \( MNO \), sum of angles is \( 180^\circ \). So angle at \( M \) is \( 180 - 76 - 41 = 63^\circ \)? Wait, no, wait. Wait, triangle \( JKL \): angle at \( K \) is \( 76^\circ \), side \( JK = 10 \), \( KL = 14 \). Triangle \( MNO \): angle at \( N = 76^\circ \), angle at \( O = 41^\circ \), sides \( ON = 70 \), \( NM = 50 \), \( OM = 75 \). Wait, first, check angles. In triangle \( JKL \), angle at \( K = 76^\circ \). Let's find angle in \( JKL \): sum of angles is \( 180 \). Let's see triangle \( MNO \): angles are \( 41^\circ \) (O), \( 76^\circ \) (N), so angle M is \( 180 - 41 - 76 = 63^\circ \). Now in triangle \( JKL \), angle at \( K = 76^\circ \). Let's check if any angles match. Wait, maybe check the sides and angles for similarity (AA or SAS). Wait, let's check the angles first. Wait, triangle \( JKL \): angle at \( K = 76^\circ \). Triangle \( MNO \): angle at \( N = 76^\circ \). Now, let's check the sides. Let's see the sides adjacent to the \( 76^\circ \) angle. In \( JKL \), sides adjacent to \( 76^\circ \) (angle K) are \( JK = 10 \) and \( KL = 14 \). In \( MNO \), sides adjacent to \( 76^\circ \) (angle N) are \( ON = 70 \) and \( NM = 50 \)? Wait, no, angle N is between sides ON and NM? Wait, ON is 70, NM is 50, and angle N is 76. In \( JKL \), angle K is 76, between JK (10) and KL (14). Let's check the ratios. \( \frac{JK}{NM} = \frac{10}{50} = \frac{1}{5} \), \( \frac{KL}{ON} = \frac{14}{70} = \frac{1}{5} \). So the sides adjacent to the equal angles (76°) are in proportion (1/5), and the included angles are equal (76°), so by SAS similarity, the triangles are similar. Wait, let's confirm the angles. Wait, in triangle \( MNO \), angle at O is 41°, angle at N is 76°, so angle at M is 63°. In triangle \( JKL \), angle at K is 76°, so let's find angle at L. Wait, maybe I made a mistake. Wait, no, let's recalculate angle in \( JKL \). Wait, triangle \( JKL \): sides JK = 10, KL = 14, angle at K = 76°. Triangle \( MNO \): sides ON = 70, NM = 50, angle at N = 76°. Wait, the sides adjacent to angle K (76°) are JK (10) and KL (14). The sides adjacent to angle N (76°) are NM (50) and ON (70)? Wait, no, angle N is between ON and NM? Wait, ON is from O to N (70), NM is from N to M (50), so angle N is between ON and NM. So the two sides forming angle N are ON (70) and NM (50). The two sides forming angle K are JK (10) and KL (14). So the ratio of JK to NM is 10/50 = 1/5, and KL to ON is 14/70 = 1/5. So the ratios are equal, and the included angles (76°) are equal. Therefore, by SAS similarity criterion, triangles JKL and MNO are similar.

Step2: Confirm SAS similarity

SAS similarity states that if two sides of one triangle are proportional to two sides of another triangle and the included angles are equal, then the triangles are similar. Here, \( \frac{JK}{NM} = \frac{10}{50} = \frac{1}{5} \), \( \frac{KL}{ON} = \frac{14}{70} = \frac{1}{5} \), and \( \angle K = \angle N = 76^\circ \). Thus, the triangles are similar by SAS.

Answer:

The triangles \( JKL \) and \( MNO \) are similar by the SAS (Side - Angle - Side) similarity criterion. The ratio of the sides adjacent to the \( 76^\circ \) angle in each triangle is \( \frac{1}{5} \) (\( \frac{10}{50}=\frac{14}{70}=\frac{1}{5} \)) and the included angles (\( 76^\circ \)) are equal.