<p>Since AB is parallel to PQ, by the alternate interior angles theorem:</p> <p>\( x = 70^\circ \)</p> <p>Since the sum of angles in a triangle is \( 180^\circ \):</p> <p>\( x + y + 105^\circ = 180^\circ \)</p> <p>Substituting the value of \( x \):</p> <p>\( 70^\circ + y + 105^\circ = 180^\circ \)</p> <p>\( y = 180^\circ - 175^\circ \)</p> <p>\( y = 5^\circ \)</p>
<p>Given that line AB is parallel to line CD, angle AEB = angle DEC as corresponding angles.</p> <p>Angle DEC = 70° (given).</p> <p>Therefore, angle AEB = 70°.</p> <p>Angle DEB = angle BEA as alternate interior angles, hence angle DEB = 28°.</p> <p>In triangle DEB, sum of angles = 180°.</p> <p>Therefore, angle EDB = 180° - 70° - 28° = 82°.</p> <p>Since angle EDB = angle EAB as alternate interior angles and line AB is parallel to line CD, angle EAB = 82° as well.</p> <p>In triangle AEB, sum of angles = 180°.</p> <p>Therefore, angle A = 180° - 70° - 82° = 28°.</p> <p>Hence, x = 28° and y = 82°.</p>
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