(a) Simplify : \(\frac{\frac{1}{3}c^{2} - \frac{2}{3}cd}{\frac{1}{2}d^{2} - \frac{1}{4}cd}\) (b) In the diagram, YPF is a straight line. < XPY = 44°, < MPF ...
Question 1 Report
(a) Simplify : \(\frac{\frac{1}{3}c^{2} - \frac{2}{3}cd}{\frac{1}{2}d^{2} - \frac{1}{4}cd}\)
(b)
In the diagram, YPF is a straight line. < XPY = 44°, < MPF = 46°, < XYP = < MFP = 90°, /XY/ = 7cm and /MP/ = 9 cm.
(i) Calculate, correct to 3 significant figures, /XM/ and /YF/ ; (ii) Find < XMP.
Answer Details
(a) Simplify the compound fraction
\[\frac{\tfrac{1}{3}c^{2} - \tfrac{2}{3}cd}{\tfrac{1}{2}d^{2} - \tfrac{1}{4}cd}\]Factorise numerator and denominator.
\[\text{Numerator} = \tfrac{1}{3}c^{2} - \tfrac{2}{3}cd = \tfrac{1}{3}c\,(c - 2d)\]\[\text{Denominator} = \tfrac{1}{2}d^{2} - \tfrac{1}{4}cd = \tfrac{1}{4}d\,(2d - c)\]Write \((c - 2d) = -(2d - c)\):
\[\frac{\tfrac{1}{3}c\,(c-2d)}{\tfrac{1}{4}d\,(2d-c)} = \frac{\tfrac{1}{3}c\,[-(2d-c)]}{\tfrac{1}{4}d\,(2d-c)} = -\frac{\tfrac{1}{3}c}{\tfrac{1}{4}d}\]\[= -\frac{1}{3}c \times \frac{4}{d} = \boxed{-\frac{4c}{3d}}\](b) Two right-angled triangles on the straight line \(YPF\)
From the diagram, \(YPF\) is a straight line, \(\angle XYP = \angle MFP = 90^\circ\), \(\angle XPY = 44^\circ\), \(\angle MPF = 46^\circ\), \(|XY| = 7\text{ cm}\) and \(|MP| = 9\text{ cm}\).
(i) Calculate \(|XM|\) and \(|YF|\).
Triangle \(XYP\) (right-angled at \(Y\)):
\[\tan 44^\circ = \frac{|XY|}{|YP|} \Rightarrow |YP| = \frac{7}{\tan 44^\circ} = \frac{7}{0.9657} = 7.249\text{ cm}\]\[\sin 44^\circ = \frac{|XY|}{|XP|} \Rightarrow |XP| = \frac{7}{\sin 44^\circ} = \frac{7}{0.6947} = 10.08\text{ cm}\]Triangle \(MPF\) (right-angled at \(F\), hypotenuse \(MP = 9\)):
\[|PF| = 9\cos 46^\circ = 9(0.6947) = 6.252\text{ cm}\]Length \(YF\): since \(Y\), \(P\), \(F\) are collinear,
\[|YF| = |YP| + |PF| = 7.249 + 6.252 = 13.5\text{ cm (3 s.f.)}\]Length \(XM\): at \(P\) the angles on the straight line give
\[\angle XPM = 180^\circ - 44^\circ - 46^\circ = 90^\circ\]so triangle \(XPM\) is right-angled at \(P\) with \(|XP| = 10.08\) and \(|PM| = 9\):
\[|XM| = \sqrt{|XP|^{2} + |PM|^{2}} = \sqrt{10.08^{2} + 9^{2}} = \sqrt{101.5 + 81} = \sqrt{182.5} = 13.5\text{ cm (3 s.f.)}\](ii) Find \(\angle XMP\). In the right-angled triangle \(XPM\) (right angle at \(P\)),
\[\tan(\angle XMP) = \frac{|XP|}{|PM|} = \frac{10.08}{9} = 1.1196\]\[\angle XMP = \tan^{-1}(1.1196) = 48.2^\circ\]Hence \(|XM| \approx 13.5\text{ cm}\), \(|YF| \approx 13.5\text{ cm}\) and \(\angle XMP \approx 48.2^\circ\).
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