Question

given: \\( \triangle abc \\) prove: \\( m\angle zab=m\angle acb + m\angle cba \\) we start with triangle \\( abc \\) and see that angle \\( zab \\) is an exterior angle created by the extension of side \\( ac \\). angles \\( zab \\) and \\( cab \\) are a linear pair by definition. we know that \\( m\angle zab + m\angle cab = 180 ^ { \circ } \\) by the we also know \\( m\angle cab + m\angle acb + m\angle cba = \\) angle addition postulate definition of complementary angles definition of vertical angles using substitution, we have \\( m\angle zab + m\angle cab \\) therefore, we conclude \\( m\angle zab = m\angle acb + m\angle cba \\) using the

Explanation:

Step1: Triangle Angle Sum Theorem

In \(\triangle ABC\), by the triangle angle - sum theorem, \(m\angle CAB + m\angle ACB+m\angle CBA=180^{\circ}\). So, \(m\angle CAB=180^{\circ}-(m\angle ACB + m\angle CBA)\).

Step2: Linear Pair of Angles

Since \(\angle ZAB\) and \(\angle CAB\) are a linear pair, \(m\angle ZAB+m\angle CAB = 180^{\circ}\). Then \(m\angle ZAB=180^{\circ}-m\angle CAB\).

Step3: Substitution

Substitute \(m\angle CAB = 180^{\circ}-(m\angle ACB + m\angle CBA)\) into \(m\angle ZAB=180^{\circ}-m\angle CAB\). We get \(m\angle ZAB=180^{\circ}-(180^{\circ}-(m\angle ACB + m\angle CBA))\).

Step4: Simplification

Simplify \(m\angle ZAB=180^{\circ}-180^{\circ}+m\angle ACB + m\angle CBA\). So, \(m\angle ZAB=m\angle ACB + m\angle CBA\).

Answer:

\(m\angle ZAB=m\angle ACB + m\angle CBA\) is proved.