Question

In the given figure, $BD=DC$ and $\angle CBD={30}^o$, find $\angle BAC$.

Answer 1

It is given that $BD=DC$

From the figure we know that

$\angle BCD=\angle CBD={30}^o$

Consider $\angle BCD$

Using the angle sum property

$\angle BCD+\angle CBD+\angle CDB={180}^o$

By substituting the values

${30}^o+{30}^o+\angle CDB={180}^o$

On further calculation

$\angle CDB={180}^o-{30}^o-{30}^o$

By subtraction

$\angle CDB={180}^o-{30}^o-{30}^o$

So we get

$\angle CDB={120}^o$

We know that the opposite angles of a cyclic angles of a cyclic quadrilateral are supplementary

It can be written as

$\angle CDB+\angle BAC={180}^o$

By substituting the values

${120}^o+\angle BAC={180}^o$

On further calculation

$\angle BAC={180}^o-{120}^o$

By subtraction

$\angle BAC={60}^o$

Therefore, $\angle BAC={60}^o$.