Question

find the values of x and y.
x = 41 (simplify your answer.)
y = (simplify your answer.)

Explanation:

Step1: Use the property of isosceles triangle

Since \(AD = BD = 5\) (marked as equal), \(\angle A=\angle ABD = 49^{\circ}\). Then for \(\triangle ABD\), using the angle - sum property of a triangle (\(\angle A+\angle ABD+\angle ADB = 180^{\circ}\)), \(\angle ADB=180-(49 + 49)=82^{\circ}\). And \(\angle CDB = 180-\angle ADB=98^{\circ}\). Also, because \(CD = BD\) (marked as equal), in \(\triangle CDB\), \(\angle C=\angle CBD\). Let \(\angle C=\angle CBD = y\). Using the angle - sum property of a triangle (\(\angle C+\angle CBD+\angle CDB = 180^{\circ}\)), \(2y+98 = 180\).

Step2: Solve for \(y\)

Subtract 98 from both sides of the equation \(2y+98 = 180\): \(2y=180 - 98=82\). Then divide both sides by 2: \(y = 41\).

Step3: Solve for \(x\)

Since \(\angle ABD = 49^{\circ}\) and \(\angle CBD=y = 41^{\circ}\), and \(\angle ABC=x+\angle ABD\). But wait, no, actually, since \(CD = BD\), \(\angle C=\angle CBD=y = 41^{\circ}\). And in \(\triangle ABD\), \(\angle ABD = 49^{\circ}\). But for \(x\), since \(AD = BD\), \(\angle A=\angle ABD = 49^{\circ}\) (base - angles of an isosceles triangle \(ABD\)).

Answer:

\(x = 49\), \(y = 41\)