In the diagram below, start overline, B, E, end overline, is perpendicular to, start overline, E, D, end overline, comma

BE

ED
, start overline, A, C, end overline, \cong, start overline, B, C, end overline, comma
AC

BC
, and m, angle, D, equals, 48, degrees, .m∠D=48

. Find m, angle, A, .m∠A.

1 answer

To solve for \( m \angle A \) given the conditions in the problem, we can follow these steps:

  1. Understand the given information:

    • Since \( \overline{BE} \) is perpendicular to \( \overline{ED} \), it means \( \angle EBD = 90^\circ \).
    • We have \( \overline{AC} \cong \overline{BC} \), which implies that triangle \( ABC \) is an isosceles triangle with \( AB = AC \).
    • It’s given that \( m \angle D = 48^\circ \).
  2. Using the properties of the triangle:

    • In triangle \( BDE \):

      • \( m \angle BDE + m \angle D + m \angle EBD = 180^\circ \)
      • We have \( m \angle EBD = 90^\circ \) and \( m \angle D = 48^\circ \).
      • Thus, we can find \( m \angle BDE \):

      \[ m \angle BDE + 48^\circ + 90^\circ = 180^\circ \] \[ m \angle BDE + 138^\circ = 180^\circ \] \[ m \angle BDE = 180^\circ - 138^\circ = 42^\circ \]

  3. Using the isosceles triangle property:

    • In triangle \( ABC \), since \( AC \cong BC \), angles \( A \) and \( C \) are equal:
    • Let \( m \angle A = m \angle C = x \).

    \[ m \angle A + m \angle C + m \angle B = 180^\circ \] \[ x + x + m \angle B = 180^\circ \] \[ 2x + m \angle B = 180^\circ \]

    We need to relate \( m \angle B \) with the angles of triangle \( BDE \).

  4. Determine \( m \angle B \):

    • Angle \( B \) in triangle \( ABC \) is supplementary to angle \( BDE \), since they share side \( \overline{BE} \).

    \[ m \angle B = 180^\circ - m \angle BDE = 180^\circ - 42^\circ = 138^\circ \]

  5. Substituting back: Plugging \( m \angle B \) into the equation for \( m \angle A \):

    \[ 2x + 138^\circ = 180^\circ \] \[ 2x = 180^\circ - 138^\circ = 42^\circ \] \[ x = \frac{42^\circ}{2} = 21^\circ \]

Therefore, \( m \angle A \) is

\[ \boxed{21^\circ}. \]