Question

if (mangle cbd = 130^{circ}), what is (mangle bad?)

Explanation:

Step1: Identify isosceles triangles

In $\triangle ABC$, since $AC = BC$, $\angle CAB=\angle CBA$. In $\triangle ABD$, since $AB = BD$, $\angle BAD=\angle BDA$.

Step2: Find $\angle ABC$

$\angle ABC = 180^{\circ}-m\angle CBD=180 - 130=50^{\circ}$.

Step3: Let $\angle BAD = x$

Let $\angle BAD=x$. Then $\angle BDA=x$. Let $\angle CAB = y$. Since $\angle ABC = 50^{\circ}$, in $\triangle ABC$, $\angle CAB=\angle CBA = y = 50^{\circ}$. In $\triangle ABD$, using the angle - sum property of a triangle ($\angle ABD = 50^{\circ}$), we have $x + x+50^{\circ}=180^{\circ}$.

Step4: Solve for $x$

Combining like terms in $2x+50^{\circ}=180^{\circ}$, we get $2x=180^{\circ}- 50^{\circ}=130^{\circ}$. Then $x = 20^{\circ}$.

Answer:

$20$