Question

determining similarity are the triangles similar? if so, state the similarity and the postulate or theorem that justifies your answer. 12. two triangles: triangle abc with angle c 110°, ac=6, cb=8; triangle ljk with angle j 110°, lj=10 2/3, jk=8

Explanation:

Step1: Identify Angles and Sides

Triangle \(ABC\) has \(\angle C = 110^\circ\), \(AC = 6\), \(CB = 8\). Triangle \(JKL\) (wait, labels: \(LJ = 10\frac{2}{3}\), \(JK = 8\), \(\angle J = 110^\circ\)? Wait, correct labels: Triangle \(ABC\): \(AC = 6\), \(CB = 8\), \(\angle C = 110^\circ\). Triangle \(LJK\): \(LJ = 10\frac{2}{3}\), \(JK = 8\), \(\angle J = 110^\circ\). Wait, check sides: \(AC = 6\), \(LJ = 10\frac{2}{3}=\frac{32}{3}\); \(CB = 8\), \(JK = 8\)? Wait no, \(CB = 8\), \(JK = 8\)? Wait, \(AC = 6\), \(LJ = \frac{32}{3}\); \(CB = 8\), \(JK = 8\)? Wait, ratio of \(AC\) to \(LJ\): \(6:\frac{32}{3}=\frac{18}{32}=\frac{9}{16}\)? No, wait maybe \(AC\) corresponds to \(JK\), \(CB\) corresponds to \(LJ\)? Wait, \(\angle C = \angle J = 110^\circ\) (given angles). Now check side ratios: \(AC = 6\), \(JK = 8\); \(CB = 8\), \(LJ = 10\frac{2}{3}=\frac{32}{3}\). Let's compute ratios: \(\frac{AC}{JK}=\frac{6}{8}=\frac{3}{4}\); \(\frac{CB}{LJ}=\frac{8}{\frac{32}{3}}=\frac{8\times3}{32}=\frac{24}{32}=\frac{3}{4}\). So two sides in proportion and included angle equal.

Step2: Apply SAS Similarity

SAS Similarity: If two sides of one triangle are proportional to two sides of another triangle and the included angles are equal, triangles are similar. Here, \(\angle C = \angle J = 110^\circ\) (included angles), and \(\frac{AC}{JK}=\frac{CB}{LJ}=\frac{3}{4}\). So triangles \(ABC \sim LJK\) by SAS Similarity.

Answer:

Yes, the triangles are similar by the SAS (Side - Angle - Side) Similarity Postulate. Triangle \(ABC \sim\) Triangle \(LJK\) (or \(ABC \sim JKL\) depending on correspondence) with \(\angle C=\angle J = 110^\circ\) and \(\frac{AC}{JK}=\frac{CB}{LJ}=\frac{3}{4}\).