Question

in the diagram, m∠acb = 55. find m∠bcd. m∠bcd = □°.

Explanation:

Step1: Identify right angle at C

From the diagram, \( \angle ACB \) and \( \angle ACD \) form a linear pair with a right angle (since \( \angle ACE \) or similar is a right angle, implying \( \angle ACB + \angle BCD = 90^\circ \)? Wait, no, looking at the diagram, \( \angle ACB \) and \( \angle BCD \) are complementary? Wait, actually, the right angle is at \( \angle AC E \) or \( \angle AC B \) and \( \angle BCD \): Wait, the diagram shows a right angle at C between A and E (or A and B? Wait, the right angle symbol is between A and C and the horizontal line. So \( \angle ACB \) is 55 degrees, and \( \angle ACB + \angle BCD = 90^\circ \)? Wait, no, the right angle is \( \angle ACE \), so \( \angle ACB \) and \( \angle BCD \): Wait, actually, since \( \angle ACB + \angle BCD = 90^\circ \)? Wait, no, the straight line is E-C-B, and A-C is perpendicular to E-C-B? Wait, the right angle symbol is at C between A and the horizontal line (E-C-B). So \( \angle ACB \) is 55 degrees, and \( \angle ACB + \angle BCD = 90^\circ \)? Wait, no, \( \angle ACD \) is a right angle? Wait, no, the right angle is \( \angle ACE \), so \( \angle ACB \) and \( \angle BCD \) are complementary because \( \angle ACB + \angle BCD = 90^\circ \) (since \( \angle ACD \) is a right angle? Wait, maybe the diagram has A-C perpendicular to E-C-B, so \( \angle ACB = 55^\circ \), then \( \angle BCD = 90^\circ - 55^\circ \).

Step2: Calculate \( m\angle BCD \)

Since \( \angle ACB + \angle BCD = 90^\circ \) (because \( AC \perp EB \), so the angle between AC and EB is 90 degrees, so \( \angle ACB \) and \( \angle BCD \) are complementary), we have:
\( m\angle BCD = 90^\circ - m\angle ACB \)
Substitute \( m\angle ACB = 55^\circ \):
\( m\angle BCD = 90 - 55 = 35^\circ \)

Answer:

35