Question

13 - 14. find x, y, and z:
13.
14.

Explanation:

Step1: Find \(x\) in problem 13

In \(\triangle ACD\), using the fact that the sum of angles in a triangle is \(180^{\circ}\).
\(x = 180^{\circ}-90^{\circ}-40^{\circ}=50^{\circ}\)

Step2: Find \(z\) in problem 13

Since \(\triangle ACD\sim\triangle CBD\) (by AA similarity, as \(\angle ADC=\angle CDB = 90^{\circ}\) and \(\angle ACD+\angle BCD = 90^{\circ}\), \(\angle B+\angle BCD=90^{\circ}\), so \(\angle A=\angle BCD\)), \(z = 40^{\circ}\)

Step3: Find \(y\) in problem 13

\(y=90^{\circ}-z = 90^{\circ}-40^{\circ}=50^{\circ}\)

Step4: Find \(x\) in problem 14

In \(\triangle ACD\), \(x = 180^{\circ}-90^{\circ}-34^{\circ}=56^{\circ}\)

Step5: Find \(z\) in problem 14

Since \(\triangle ACD\sim\triangle CBD\) (by AA similarity, \(\angle ADC=\angle CDB = 90^{\circ}\) and \(\angle ACD+\angle BCD = 90^{\circ}\), \(\angle B+\angle BCD=90^{\circ}\), so \(\angle A=\angle BCD\)), \(z = 34^{\circ}\)

Step6: Find \(y\) in problem 14

\(y=90^{\circ}-z=90^{\circ}-34^{\circ}=56^{\circ}\)

Answer:

For problem 13: \(x = 50^{\circ}\), \(y = 50^{\circ}\), \(z = 40^{\circ}\)
For problem 14: \(x = 56^{\circ}\), \(y = 56^{\circ}\), \(z = 34^{\circ}\)