Question

ABCD s a rhombus with diagonals that intersect at point E.
EB=15, AB=25, and m ABD is 33

what is the measure of BDA

Answer 1

To find the measure of angle \( BDA \) in rhombus \( ABCD \), we begin with the given information:

- \( EB = 15 \)
- \( AB = 25 \)
- \( m \angle ABD = 33^\circ \)

Since \( ABCD \) is a rhombus, the diagonals \( AC \) and \( BD \) bisect each other at point \( E \). So, \( AE = EC \) and \( BE = ED \). Given \( EB = 15 \), we find that \( ED = 15 \) as well, which means \( BD = BE + ED = 15 + 15 = 30 \).

Using one of the properties of rhombuses, the diagonals are perpendicular bisectors of each other. Thus, \( \angle AEB = 90^\circ \).

Next, we can find angle \( ABD \) using triangle \( ABE \):
- We know \( \angle ABE = 33^\circ \).
- Since \( \angle AEB = 90^\circ \), we can find angle \( BAE \):

\[
m \angle ABE + m \angle AEB + m \angle BAE = 180^\circ
\]
\[
33^\circ + 90^\circ + m \angle BAE = 180^\circ
\]
\[
m \angle BAE = 180^\circ - 33^\circ - 90^\circ
\]
\[
m \angle BAE = 57^\circ
\]

Now, consider triangle \( ADB \). Since \( ADB \) and \( ABD \) share \( AB \) and \( AD \) (both are sides of the rhombus and equal), and since \( angles \angle AEB \) and \( \angle ADB \) are supplementary to angle \( BDA \).

In triangle \( ADB \),

\[
m \angle ADB = m \angle ABE = 33^\circ
\]
Using the sum of angles in triangle \( ADB \):

\[
m \angle ADB + m \angle ABD + m \angle BDA = 180^\circ
\]

Substitute the known values:

\[
33^\circ + 33^\circ + m \angle BDA = 180^\circ
\]

Combine and solve for \( m \angle BDA \):

\[
66^\circ + m \angle BDA = 180^\circ
\]
\[
m \angle BDA = 180^\circ - 66^\circ
\]
\[
m \angle BDA = 114^\circ
\]

Thus, the measure of angle \( BDA \) is:

\[
\boxed{114^\circ}
\]

Answer 2

that's not right