Question

triangle abc, shown in the diagram below, is an isosceles triangle. if the measure of ∠acd is 140°, what is the measure of ∠cab? 140° 100° 40° 80°

Explanation:

Step1: Find adjacent - interior angle

$\angle ACB=180^{\circ}-\angle ACD = 180 - 140=40^{\circ}$

Step2: Use isosceles - triangle property

In $\triangle ABC$, since it is isosceles with $AC = AB$, $\angle ABC=\angle ACB = 40^{\circ}$

Step3: Calculate $\angle CAB$

$\angle CAB=180^{\circ}-\angle ABC-\angle ACB=180 - 40 - 40 = 100^{\circ}$ (This is wrong above, let's correct from step 2). In $\triangle ABC$, since it is isosceles with $AC = AB$, $\angle ABC=\angle ACB$. And $\angle ACD$ is an exterior angle of $\triangle ABC$. By the exterior - angle theorem, $\angle ACD=\angle CAB+\angle ABC$. Since $\angle ABC=\angle ACB$, and $\angle ACD = 140^{\circ}$, and $\angle CAB=\angle ACD-\angle ABC$. Also, $\angle ABC=\angle ACB$. Let $\angle CAB = x$, then $\angle ABC=\angle ACB = 180 - 140=40^{\circ}$. So $\angle CAB = 40^{\circ}$

Answer:

C. $40^{\circ}$