Find the third angle of a triangle (using a parallel line) when two of the angles are \(35°\) and \(71°\).
Answer:
Let \(ABC\) be the given triangle, where \(\angle B = 71°\) and \(\angle C = 35°\).
Let us suppose we construct a line \(XY\) parallel to \(BC\) through the vertex \(A\).
\(\angle BAX =\) \(°\) [Since ]
\(\angle CAY =\) \(°\) [Since ]
\(\angle BAC =\) \(°\) [By ]
Therefore, the third angle of the triangle \(ABC\) is \(°\).