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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} )
Step1: Recall the triangle angle - sum property
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: Use the given inequality \(m\angle Z>m\angle X + m\angle Y\)
Substitute \(m\angle X + m\angle Y = 180^{\circ}-m\angle Z\) into \(m\angle Z>m\angle X + m\angle Y\). We get \(m\angle Z>180^{\circ}-m\angle Z\).
Step3: Solve the inequality for \(m\angle Z\)
Add \(m\angle Z\) to both sides: \(m\angle Z+m\angle Z>180^{\circ}\), \(2m\angle Z>180^{\circ}\), \(m\angle Z > 90^{\circ}\).
Step4: Analyze \(m\angle X + m\angle Y\)
Since \(m\angle X + m\angle Y+m\angle Z=180^{\circ}\) and \(m\angle Z>90^{\circ}\), then \(m\angle X + m\angle Y=180^{\circ}-m\angle Z<90^{\circ}\).
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\(m\angle X + m\angle Y < 90^{\circ}\) (the fourth option)