Question

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

Explanation:

Step1: Find angle BAC

In right - triangle ABC with $\angle C = 90^{\circ}$ and $\angle ABC=30^{\circ}$, using the angle - sum property of a triangle ($\angle BAC+\angle ABC+\angle C = 180^{\circ}$), we have $\angle BAC=180^{\circ}-\angle C-\angle ABC=180^{\circ}-90^{\circ}-30^{\circ}=60^{\circ}$.

Step2: Determine the measure of angle BAL

Since AL is the angle - bisector of $\angle BAC$, then $\angle BAL=\frac{1}{2}\angle BAC=\frac{1}{2}\times60^{\circ}=30^{\circ}$.

Step3: Analyze triangle ABL

In $\triangle ABL$, $\angle BAL = 30^{\circ}$ and $\angle ABC = 30^{\circ}$, so $\triangle ABL$ is an isosceles triangle, and $AL = LB = 18$ m.

Step4: Find CL in right - triangle ACL

In right - triangle ACL, $\angle CAL = 30^{\circ}$, and we know that in a right - triangle, if the angle is $30^{\circ}$, the side opposite the $30^{\circ}$ angle is half of the hypotenuse. Here, CL is the side opposite the $30^{\circ}$ angle $\angle CAL$ and AL is the hypotenuse. So $CL=\frac{1}{2}AL$.
Since $AL = 18$ m, then $CL = 9$ m.

Answer:

9 m