if ABC was bisected by BD, what is the mABC (BDC is 33 degrees)

Answer 1

If \( BD \) bisects \( \angle ABC \), it means that \( \angle ABD \) and \( \angle DBC \) are equal.

Given that \( m\angle BDC = 33^\circ \), we can use the fact that the angles around point \( B \) add up to \( 180^\circ \):

\[ m\angle ABC + m\angle BDC = 180^\circ \]

Since \( m\angle ABC \) is composed of \( \angle ABD + \angle DBC \), and each of those is half of \( m\angle ABC \) (because \( BD \) bisects it), we can express the relationship as follows:

Let \( m\angle ABC = x \). Then:

\[ x + 33^\circ = 180^\circ \]

Solving for \( x \):

\[ x = 180^\circ - 33^\circ \] \[ x = 147^\circ \]

Thus, \( m\angle ABC = 147^\circ \).

Answer 2

the only options is 66, 132, 16.5 and 33

Answer 3

Given that \( BD \) bisects \( \angle ABC \) and that \( m\angle BDC = 33^\circ \), we can determine \( m\angle ABC \) based on the bisection.

Since \( BD \) is the angle bisector, we have:

\[ m\angle ABD = m\angle DBC = \frac{m\angle ABC}{2} \]

Now, since \( \triangle BDC \) is formed by \( \angle BDC \), and given that \( m\angle BDC = 33^\circ \), we can find that:

\[ m\angle ABC + m\angle BDC = 180^\circ \]

Since \( BD \) bisects \( \angle ABC \), let's express \( m\angle ABC \) as \( x \):

\[ x + 33^\circ = 180^\circ \]

Solving for \( x \):

\[ x = 180^\circ - 33^\circ \] \[ x = 147^\circ \]

However, \( 147^\circ \) is not one of the available options. Given the choices \( 66^\circ, 132^\circ, 16.5^\circ, \) and \( 33^\circ \), we can find half of \( m\angle ABC \) since \( BD \) is the bisector:

\[ m\angle ABC = 2 \times m\angle ABD = 2 \times m\angle DBC \]

From \( m\angle DBC = 33^\circ \):

So, if \( m\angle DBC = 33^\circ \):

\[ m\angle ABC = 2 \times 33^\circ = 66^\circ \]

Thus, the correct answer among the options is:

\[ \boxed{66^\circ} \]