Question

solve the following problems. given: △abc, m∠c = 90°, m∠abc = 30°, $overline{al}$ -∠ bisector, lb = 18 m. find: cl. answer:

Explanation:

Step1: Find angle BAC

In right - triangle ABC, $\angle C = 90^{\circ}$, $\angle ABC=30^{\circ}$, so $\angle BAC = 60^{\circ}$. Since AL is the angle - bisector, $\angle LAC=\angle LAB = 30^{\circ}$.

Step2: Use angle - side relationship

In $\triangle ABL$, $\angle LAB=\angle ABC = 30^{\circ}$, so $AL = LB=18$ m.

Step3: Find CL in right - triangle ACL

In right - triangle ACL, $\sin\angle LAC=\frac{CL}{AL}$. Since $\angle LAC = 30^{\circ}$ and $AL = 18$ m, and $\sin30^{\circ}=\frac{1}{2}$, then $CL=\frac{1}{2}AL$.

Answer:

9 m