Find the third angle of a triangle using a parallel line when two of the angles are \(45°\) and \(77°\).
Answer:
Let \(ABC\) be the given triangle, where \(\angle B = 77°\) and \(\angle C = 45°\).
Let us suppose we construct a line \(XY\) parallel to the base line \(BC\) through the vertex \(A\).
\(\angle BAX =\) \(°\) [Since ]
\(\angle CAY =\) \(°\) [Since ]
\(\angle BAC =\) \(°\) [By ]
Therefore, the third angle of the triangle \(ABC\) is \(°\).