Question

8 the diagonals of rectangle wxyz intersect at v. given that $mangle vxy=23^{circ}$, find $mangle wxz$ and $mangle wyx$. $mangle wxz = square^{circ}$ $mangle wyx = square^{circ}$

Explanation:

Step1: Recall rectangle diagonal properties

In rectangle \(WXYZ\), diagonals bisect each other, so \(VX = VY\). Thus, \(\triangle VXY\) is isosceles, so \(m\angle VXY = m\angle VYX = 23^\circ\).

Step2: Find \(m\angle WXZ\)

All angles in a rectangle are \(90^\circ\), so \(m\angle WXY = 90^\circ\).
\(m\angle WXZ = m\angle WXY - m\angle VXY\)
\(m\angle WXZ = 90^\circ - 23^\circ = 67^\circ\)

Step3: Find \(m\angle WYX\)

First, calculate \(m\angle XVY\) using triangle angle sum:
\(m\angle XVY = 180^\circ - 23^\circ - 23^\circ = 134^\circ\)
Vertical angles are equal, so \(m\angle WVY = 134^\circ\). Diagonals of a rectangle are equal and bisect each other, so \(WV = VY\), making \(\triangle WVY\) isosceles.
Let \(m\angle WYX = x\), then:
\(2x + 134^\circ = 180^\circ\)
\(2x = 180^\circ - 134^\circ = 46^\circ\)
\(x = \frac{46^\circ}{2} = 23^\circ\)

Answer:

\(m\angle WXZ = 67^\circ\)
\(m\angle WYX = 23^\circ\)