Triangles [NEW] — Maths STD 9 — Question

2. 80º

Solution:

AC || BD

And, AB is transverse to these parallal lines

So $\angle\text{CAB}=\angle\text{ABD}$ (Alternate angles)

$\Rightarrow\angle\text{ABD}=45^\circ$

Now In $\triangle\text{BOD}$

$\angle\text{BOD}+\angle\text{ODB}+\angle\text{DBA}=180^\circ$

$\angle\text{DBA}=\angle\text{ABD}=45^\circ,\angle\text{ODB}=55^\circ$

So $\angle\text{BOD}=180^\circ-45^\circ-55^\circ$

$=80^\circ$