Question

in triangle xyz, ( mangle z>mangle x + mangle y ). which must be true about ( \triangle xyz )?
( mangle x + mangle z<90^{circ} )
( mangle y>90^{circ} )
( angle x ) and ( angle y ) are complementary
( mangle x + mangle y<90^{circ} )

Explanation:

Step1: Use the triangle - angle sum theorem

In any triangle \( \triangle XYZ\), \(m\angle X + m\angle Y+m\angle Z=180^{\circ}\), so \(m\angle X + m\angle Y = 180^{\circ}-m\angle Z\).

Step2: Substitute into the given inequality

Given \(m\angle Z>m\angle X + m\angle Y\). Substitute \(m\angle X + m\angle Y = 180^{\circ}-m\angle Z\) into the inequality: \(m\angle Z>180^{\circ}-m\angle Z\).

Step3: Solve the inequality for \(m\angle Z\)

Add \(m\angle Z\) to both sides: \(2m\angle Z>180^{\circ}\). Then divide both sides by 2: \(m\angle Z > 90^{\circ}\).
Since \(m\angle X + m\angle Y+m\angle Z=180^{\circ}\), if \(m\angle Z>90^{\circ}\), then \(m\angle X + m\angle Y=180^{\circ}-m\angle Z<90^{\circ}\).

Answer:

\(m\angle X + m\angle Y<90^{\circ}\)