75. In quadrilateral $A B C D$, it is given that $\widehat{D A B}=150^{\circ}$, $\widehat{D A C}+\widehat{A B D}=120^{\circ}, \widehat{D B C}-\widehat{A B \bar{D}}=60^{\circ}$. Find $\widehat{B D C}$.
75. Let $\widehat{A B D}=\alpha, \widehat{B D C}=\varphi$. By the condition, $\widehat{D A C}=$ $=120^{\circ}-\alpha, \widehat{B A C}=30^{\circ}+\alpha, \widehat{A D B}=30^{\circ}-\alpha, \widehat{D B C}=60^{\circ}+\alpha$. Using the Law of Sines for triangles $A B C, B C D, A C D$, we get

$$
\begin{aligned}
& \frac{|B C|}{|A C|}=\frac{\sin \left(30^{\circ}+\alpha\right)}{\sin \left(60^{\circ}+2 \alpha\right)}=\frac{1}{2 \cos \left(30^{\circ}+\alpha\right)}, \\
& \frac{|D C|}{|B C|}=\frac{\sin \left(60^{\circ}+\alpha\right)}{\sin \varphi}, \\
& \frac{|A C|}{|D C|}=\frac{\sin \left(30^{\circ}-\alpha+\varphi\right)}{\sin \left(120^{\circ}-\alpha\right)} .
\end{aligned}
$$

Multiplying these equations, we have

$\sin \left(30^{\circ}-\alpha+\varphi\right)=\sin \left(30^{\circ}+\alpha+\varphi\right)-\sin \left(30^{\circ}+\alpha-\varphi\right) \Rightarrow$

$\Rightarrow \sin \left(30^{\circ}-\alpha+\varphi\right)+\sin \left(30^{\circ}+\alpha-\varphi\right)=\sin \left(30^{\circ}+\alpha+\varphi\right) \Rightarrow$

$\Rightarrow \cos (\varphi-\alpha)=\sin \left(30^{\circ}+\alpha+\varphi\right) \Rightarrow$

$\Rightarrow \sin \left(90^{\circ}-\varphi+\alpha\right)-\sin \left(30^{\circ}+\alpha+\varphi\right)=$

$$
=2 \cos \left(60^{\circ}+\alpha\right) \sin \left(30^{\circ}-\varphi\right)=0 ;
$$

thus, $\varphi=30^{\circ}$.