In the given figure find angle x. I approached it these ways-
1)Constructing parallel lines and then using similarity
2)Trying exterior angle property
3)Constructing perpendiculars
I am not able to understand the suitable construction.

 A: Hint:
Draw $\angle ABE = 15^0$. Then what is $\angle BED$? Can you show $\triangle BDE, \triangle BEA \,$ and $\triangle BEC$ are isosceles? Also what is $\angle CED$? From there, it is straightforward to show $\angle C = 75^0$.
Just an observation - $E$ is the circumcenter of $\triangle ABC$.

A: 
This is almost similar to Math Lover's solution...
Draw $AH$ perpendicular to $BD$. In the right triangle AHD:
$$\widehat{HAD}=90^o-60^o=30^o \quad \Rightarrow \quad HD=\frac12AD=1 $$
Draw HC and observe that triangle $DHC$ is isosceles:
$$ HD=1=DC \quad \Rightarrow \quad \widehat{DHC} = \widehat{DCH} = \frac12(180^o-120^o)=30^o$$
Now observe that triangle $HAC$ is also isosceles:
$$\widehat{HAC}=\widehat{HCA}=30^0 \quad \Rightarrow \quad HA=HC \qquad(1)$$
In triangle $DBC$ note that $\widehat{DBC}=180^o-(120^o+45^o)=15^o$.
Now, observe that triangle $HBC$ is isosceles too:
$$\widehat{HCB} = \widehat{ACB}-\widehat{ACH}=45^o-30^o=15^o $$
$$\widehat{HCB}=\widehat{HBC}=15^o \quad \Rightarrow \quad HC=HB \qquad(2) $$
From (1) and (2) we conclude that $HA=HB$ and that the right triangle $AHB$ is isosceles, which means $\widehat{HAB}=45^o$. Now we can say that
$$\widehat{BAC}=\widehat{BAH}+\widehat{HAC}=45^o+30^o=75^o $$
