To solve the problem, we use trigonometric identities and the Law of Sines in the triangles formed by point D inside $\triangle ABC$.

Step 1: Calculate Known Angles

• $\angle ABC = \angle DBA + \angle DBC = 20^\circ + 40^\circ = 60^\circ$
• Let $\angle ACD = x$ (target angle). Then $\angle ACB = x + 30^\circ$ (since $\angle DCB=30^\circ$).
• $\angle BAC = \alpha$, so $\alpha + (x+30^\circ) + 60^\circ = 180^\circ \implies \alpha = 90^\circ - x$.
• $\angle DAC = \alpha - 10^\circ = 80^\circ - x$.

Step 2: Apply Law of Sines

Using the Law of Sines in $\triangle ABD$, $\triangle BDC$, and $\triangle ADC$:

• In $\triangle ABD$: $\frac{AD}{\sin 20^\circ} = \frac{BD}{\sin 10^\circ} \implies AD = BD \cdot \frac{\sin 20^\circ}{\sin 10^\circ}$.
• In $\triangle BDC$: $\frac{CD}{\sin 40^\circ} = \frac{BD}{\sin 30^\circ} \implies CD = BD \cdot 2\sin 40^\circ$.
• In $\triangle ADC$: $\frac{AD}{\sin x} = \frac{CD}{\sin (80^\circ - x)}$.

Step 3: Simplify the Equation

Substitute $AD$ and $CD$ into the equation for $\triangle ADC$:
$\frac{\sin 20^\circ}{\sin 10^\circ \sin x} = \frac{2\sin 40^\circ}{\sin (80^\circ - x)}$.

Since $\sin 20^\circ = 2\sin 10^\circ \cos 10^\circ$, this simplifies to:
$\frac{2\cos 10^\circ}{\sin x} = \frac{2\sin 40^\circ}{\sin (80^\circ - x)}$.

Cancel 2 and cross-multiply:
$\cos 10^\circ \sin (80^\circ - x) = \sin 40^\circ \sin x$.

For $x=50^\circ$: $\sin(30^\circ) \cos 10^\circ = \sin 40^\circ \sin 50^\circ$ (since $\sin 50^\circ = \cos 40^\circ$, $\sin 40^\circ \cos 40^\circ = \frac{1}{2}\sin 80^\circ = \frac{1}{2}\cos 10^\circ$).

This holds true, so $x=50^\circ$.

Answer: $\boxed{50}$

作者声明:本文包含人工智能生成内容。