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To solve this problem, we need to find angle \( \angle BCA \) in triangle \( \triangle ABC \). Given: - \( \angle BAC = 31^\circ \) - \( \angle BDC = 75^\circ \) Since \( D \) lies on \( AC \), \( \angle BDC \) is an external angle for triangle \( \triangle ABC \). The external angle theorem states that \( \angle BDC = \angle BAC + \angle BCA \). Thus, we have: \[ \angle BDC = \angle BAC + \angle BCA \] Substitute the known values: \[ 75^\circ = 31^\circ + \angle BCA \] Solve for \( \angle BCA \): \[ \angle BCA = 75^\circ - 31^\circ = 44^\circ \] Thus, \( \angle BCA = 44^\circ \).