(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^\circ, < MPF = 46^\circ, < XYP = < MFP = 90^\circ, /XY/ = 7cm and /MP/ = 9 cm.
(i) Calculate, correct to 3 significant figures, /XM/ and /YF/ ; (ii) Find < XMP.
Verified Correct Answer: .
Step-by-Step Mathematical Solution
(a) \(\frac{\frac{1}{3}c^{2} - \frac{2}{3}cd}{\frac{1}{2}d^{2} - \frac{1}{4}cd}\)
= \(\frac{\frac{1}{3}c(c - 2d)}{\frac{1}{4}d(2d - c)}\)
= \(\frac{\frac{1}{3}c (c - 2d)}{-\frac{1}{4}d (c - 2d)}\)
= \(-\frac{4c}{3d}\)
(b)(i) In \(\Delta XYP\),
\(\frac{7}{XP} = \sin 44^\circ\)
\(XP = \frac{7}{\sin 44}\)
= \(10.08 cm\)
< XPM + 44^\circ + 46^\circ = 180^\circ
< XPM = 90^\circ.
In \(\Delta XPM\),
\(XM^{2} = XP^{2} + PM^{2}\)
\(XM^{2} = (10.08)^{2} + 9^{2}\)
= \(101.6064 + 81 = 182.6064\)
\(XM = \sqrt{182.6064}\)
= \(15.513 cm \approxeq 13.5 cm\)
In \(\Delta XYP\),
\(\frac{7}{YP} = \tan 44\)
\(YP = \frac{7}{\tan 44}\)
\(YP = 7.249 cm \approxeq 7.2 cm\)
In \(\Delta MPF\),
\(\frac{PF}{9} = \cos 46\)
\(PF = 9 \cos 46 = 6.2519 cm\)
\(YF = YP + PF = 7.249 + 6.2519\)
= \(13.5009 cm \approxeq 13.5 cm\)
(ii)
\(\frac{10.1}{9} = \tan \theta\)
\(\tan \theta = 1.122\)
\(\theta = \tan^{-1} (1.122)\)
= 48.296^\circ
Source: Myschool Past Questions Archive (Record #53286)