$24^{\\circ}$\n$30^{\\circ}$\n?

Answer

Explanation:

Step1: Label the triangle vertices and points

Let the triangle be $\triangle ABC$ with $A$ at the top, $B$ at the bottom left, and $C$ at the bottom right. Let $D$ be the point on $BC$ where the internal line segment meets.

Step2: Identify given values and congruences

Given $\angle ABD = 24^{\circ}$, $\angle BAD = 30^{\circ}$, and side lengths $AB = DC$. Let $AB = DC = c$.

Step3: Calculate the third angle in $\triangle ABD$

Using the sum of angles in a triangle: $\angle ADB = 180^{\circ} - (24^{\circ} + 30^{\circ}) = 126^{\circ}$

Step4: Find the supplementary angle $\angle ADC$

$\angle ADC = 180^{\circ} - 126^{\circ} = 54^{\circ}$

Step5: Apply the Law of Sines in $\triangle ABD$

Find length $AD$ in terms of $c$: $\frac{AD}{\sin(24^{\circ})} = \frac{c}{\sin(126^{\circ})} \implies AD = \frac{c \sin(24^{\circ})}{\sin(126^{\circ})}$

Step6: Apply the Law of Sines in $\triangle ADC$

Let $\angle ACD = x$. Then $\angle DAC = 180^{\circ} - (54^{\circ} + x) = 126^{\circ} - x$. $\frac{DC}{\sin(126^{\circ} - x)} = \frac{AD}{\sin(x)} \implies \frac{c}{\sin(126^{\circ} - x)} = \frac{c \sin(24^{\circ})}{\sin(126^{\circ}) \sin(x)}$

Step7: Simplify the trigonometric equation

$\sin(126^{\circ}) \sin(x) = \sin(24^{\circ}) \sin(126^{\circ} - x)$ $\sin(54^{\circ}) \sin(x) = \sin(24^{\circ}) (\sin(126^{\circ})\cos(x) - \cos(126^{\circ})\sin(x))$

Step8: Solve for $x$

$\sin(54^{\circ}) \sin(x) = \sin(24^{\circ}) \sin(54^{\circ}) \cos(x) + \sin(24^{\circ}) \cos(54^{\circ}) \sin(x)$ Divide by $\sin(x) \sin(54^{\circ})$: $1 = \sin(24^{\circ}) \cot(x) + \sin(24^{\circ}) \cot(54^{\circ})$ $\cot(x) = \frac{1 - \sin(24^{\circ}) \cot(54^{\circ})}{\sin(24^{\circ})} = \frac{1}{\sin(24^{\circ})} - \cot(54^{\circ})$ Using identity $\frac{1}{\sin(24^{\circ})} - \frac{\cos(54^{\circ})}{\sin(54^{\circ})} = \cot(12^{\circ})$, we find $x = 12^{\circ}$.

Answer:

$12^{\circ}$