(a) In the diagram, MN || ST, NP || QT and < STQ = 70°. Find x. (b) In the diagram above, AC is a straight line, |BC| = |BD|, \(\stackrel\frown{BCD} = 50°\)...

Answer Details

(a) Find \(x\).

From the diagram, \(MN\parallel ST\) and \(NP\parallel QT\), with \(\angle STQ=70^{\circ}\) at \(T\), and \(x\) is the angle at \(N\) between \(NM\) and \(NP\).

Extend \(QT\) to cut \(MN\) at a point \(W\). Because \(MN\parallel ST\) and \(QT\) is a transversal, corresponding angles give

\[\angle MWQ=\angle STQ=70^{\circ}\]

Now \(MN\) is a transversal of the parallel lines \(NP\) and \(QT\). The ray \(NP\) points to the same side of \(MN\) as \(P\) (above), while \(WQ\) points to the opposite side (below), so \(x\) and \(\angle MWQ\) are co-interior (allied) angles:

\[x+\angle MWQ=180^{\circ}\]\[x=180^{\circ}-70^{\circ}=\mathbf{110^{\circ}}\]

(b) Find \(\angle BDA\).

From the diagram \(A,B,C\) lie on the straight line \(AC\), with \(D\) above it, \(|BC|=|BD|\), \(\angle BCD=50^{\circ}\) and \(\angle BAD=55^{\circ}\).

Triangle \(BCD\) is isosceles with \(|BC|=|BD|\), so the base angles are equal:

\[\angle BDC=\angle BCD=50^{\circ}\]\[\angle DBC=180^{\circ}-50^{\circ}-50^{\circ}=80^{\circ}\]

Since \(A,B,C\) are collinear, \(\angle DBA\) and \(\angle DBC\) are angles on a straight line at \(B\):

\[\angle DBA=180^{\circ}-80^{\circ}=100^{\circ}\]

In triangle \(ABD\):

\[\angle BDA=180^{\circ}-\angle BAD-\angle DBA=180^{\circ}-55^{\circ}-100^{\circ}=\mathbf{25^{\circ}}\]