Given:
BAC = 33 degrees
BDC = 35 degrees
Solution:
From the properties of an isosceles triangle:
The base angles of an isosceles triangle are equal. Hence from triangle BDC, we have:
[tex]\angle\text{BDC = }\angle\text{BCD = 35}^0[/tex]
We can obtain angle DBC using the theorem that the sum of angles in a triangle is 180 degrees:
[tex]\begin{gathered} \angle DBC=180^0-35^0-35^0 \\ =110^0 \end{gathered}[/tex]
To find angle ABD, we use the theorem of congruency. i.e
[tex]\Delta\text{ ABD }\cong\text{ }\Delta\text{ ABC}[/tex]
Hence,
[tex]\angle\text{ ABD = }\angle\text{ ABC}[/tex]
Since the angles ABD, ABC and DBC lie at a point, we have:
[tex]\begin{gathered} Let\text{ }\angle\text{ ABD = x} \\ x+x+110^0=360^0 \\ 2x=250^0 \\ x=125^0 \end{gathered}[/tex]
Answer : angle ABD = 125 degrees
