Congruence and proofs
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\(PQ\) is parallel to \(RS\), \(\angle QPT = 40^\circ\), \(\angle QTS = 110^\circ\). Find, giving reasons, the size of \(\angle TRS\).
\(\begin{align}
&\angle RST = \angle QPT = 40^\circ \\
&\text{(alternate angles = in parallel lines RS, PQ)} \\
&\angle TRS + \angle RST = \angle QTS \\
&\text{(exterior }\angle \text{ of }\Delta TRS \text{ is }= \text{ to the sum of the two opposite interior } \angle\text{'s)}\\
&\therefore \angle TRS + 40^\circ = 110^\circ\\
&\therefore \angle TRS = 70^\circ
\end{align}\)
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