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2003 WAEC Mathematics Theory (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...

Mathematics
WAEC 2003

(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.

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Explanation

(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 \(\Delta XYP\),

\(\frac{7}{XP} = \sin 44°\)

\(XP = \frac{7}{\sin 44}\)

= \(10.08 cm\)

< XPM + 44° + 46° = 180°

< XPM = 90°.

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° 


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