Answer:
[tex]\angle\text{BDC}=44^\circ[/tex]
Step-by-step explanation:
First, let's work out for angle ABD:
In quadrilateral ABDE,
[tex]\angle \text{A}+\angle \text{E}+\angle \text{BDE}+\angle \text{ABD}=360^\circ\ \ \ [\text{Sum of angles of quadrilateral is 360}^\circ.]\\\\\text{or, }120^\circ+102^\circ+54^\circ+\angle \text{ABD}=360^\circ\\\\\text{or, }\angle \text{ABD}=84^\circ[/tex]
Now in triangle BCD,
[tex]\angle \text{BDC}+\angle \text{C}=\angle \text{ABD}\ \ \ [\text{An exterior angle of a triangle is equal to the sum of}\\[/tex]
[tex]\text{the opposite interior angles.}][/tex]
[tex]\text{or, }\angle\text{BDC}+40^\circ=84^\circ\\\therefore\ \angle\text{BDC}=44^\circ[/tex]
