Question

find the values of x and y.
x = \square (simplify your answer.)

Explanation:

Step1: Identify Triangle Type

The triangle \( \triangle ABC \) has \( AB = BC \) (marked with equal ticks), so it's isosceles with \( \angle A=\angle C \)? Wait, no, \( AB \) and \( BC \) are equal? Wait, the marks: \( AB \) and \( BC \) have ticks? Wait, actually, \( AD = 2 \), \( D \) is on \( AC \), and \( BD \) is a segment. Wait, \( \triangle ABD \) and \( \triangle CBD \)? Wait, no, the triangle \( \triangle ABC \): \( AB = BC \) (since two sides have ticks), so \( \triangle ABC \) is isosceles with \( AB = BC \). Wait, \( \angle A = 33^\circ \), so \( \angle C = 33^\circ \)? No, wait, \( AB = BC \), so the base is \( AC \), so the base angles are \( \angle A \) and \( \angle C \)? Wait, no, in a triangle, equal sides have equal opposite angles. So if \( AB = BC \), then \( \angle A=\angle C \). But \( \angle A = 33^\circ \), so \( \angle C = 33^\circ \). Then \( BD \) is the angle bisector? Wait, no, \( BD \) is also a median? Wait, \( AD = 2 \), so \( AC = AD + DC = 2 + y \), but if \( AB = BC \) and \( BD \) is perpendicular? Wait, maybe \( BD \) is the altitude, median, and angle bisector in an isosceles triangle. Wait, the triangle \( \triangle ABC \) with \( AB = BC \), so it's isosceles with \( AB = BC \), so vertex at \( B \). Then \( \angle ABC = 180^\circ - 2\times33^\circ = 114^\circ \). Then \( BD \) splits \( \angle ABC \) into two equal angles? Wait, no, the marks on \( AB \) and \( BC \) are equal, and \( BD \) is a segment with \( AD = 2 \), \( DC = y \), and \( \angle ABD = \angle CBD = x \). Wait, maybe \( \triangle ABC \) is isosceles with \( AB = BC \), so \( \angle A = \angle C = 33^\circ \), then \( \angle ABC = 180 - 33 - 33 = 114^\circ \). Then \( BD \) is the angle bisector, so \( x = \frac{114^\circ}{2} = 57^\circ \)? Wait, no, wait, maybe \( AB = BC \), so \( \angle A = \angle C \), but \( \angle A = 33^\circ \), so \( \angle C = 33^\circ \), then \( \angle ABC = 180 - 33 - 33 = 114^\circ \), and \( BD \) is the angle bisector, so \( x = \frac{114}{2} = 57 \). Wait, but let's check again.

Wait, the triangle: \( AB = BC \) (marked with ticks), so \( AB = BC \), so \( \angle A = \angle C = 33^\circ \). Then \( \angle ABC = 180 - 33 - 33 = 114^\circ \). Since \( BD \) is a line from \( B \) to \( AC \), and if \( AB = BC \), then \( BD \) is the angle bisector, median, and altitude. So \( AD = DC \), so \( y = 2 \), but we need \( x \). Wait, \( \angle ABC = 114^\circ \), so \( x = \frac{114}{2} = 57^\circ \). So \( x = 57 \).

Step2: Calculate \( x \)

In \( \triangle ABC \), \( AB = BC \) (isosceles), so \( \angle A = \angle C = 33^\circ \).
Sum of angles: \( \angle A + \angle B + \angle C = 180^\circ \)
\( 33^\circ + \angle ABC + 33^\circ = 180^\circ \)
\( \angle ABC = 180 - 66 = 114^\circ \)

Since \( BD \) bisects \( \angle ABC \) (as \( AB = BC \) and \( BD \) is a median/altitude), \( x = \frac{\angle ABC}{2} = \frac{114^\circ}{2} = 57^\circ \).

Answer:

\( x = 57 \)