(a) Without using tables or calculator, simplify : \(\frac{0.6 \times 32 \times 0.004}{1.2 \times 0.008 \times 0.16}\), leaving the answer in standard form (scientific notation).
(b)
In the diagram, \(\overline{EF}\) is parallel to \(\overline{GH}\). If \(< AEF = 3x^\circ, < ABC = 120^\circ\) and \(< CHG = 7x^\circ\), find the value of \(< GHB\).
Verified Correct Answer: .
Step-by-Step Mathematical Solution
(a) \(\frac{0.6 \times 32 \times 0.004}{1.2 \times 0.008 \times 0.16}\)
= \(\frac{6 \times 10^{-1} \times 32 \times 4 \times 10^{-3}}{12 \times 10^{-1} \times 8 \times 10^{-3} \times 16 \times 10^{-2}}\)
= \(\frac{4}{2 \times 4 \times 10^{-2}}\)
= \(\frac{1}{2} \times \frac{1}{10^{-2}\)
= \(0.5 \times 10^{2} \)
= \(5 \times 10^{1}\)
(b)
In the diagram above, \(< GHB = 180 - 7x^\circ\) (angles on a straight line)
\(< HBN = 180 - 7x^\circ\) (alternate angles)
\(< EBN = 3x^\circ\) (corresponding angles)
Then \(3x^\circ + (180 - 7x)^\circ = 120^\circ\)
\(\implies 3x^\circ - 7x^\circ + 180^\circ = 120^\circ\)
\(\implies 180^\circ - 120^\circ = 7x^\circ - 3x^\circ\)
\(60^\circ = 4x^\circ\)
\(x = 15^\circ\)
Hence, \(< GHB = 180^\circ - (7 \times 15^\circ) = 180^\circ - 105^\circ\)
= \(75^\circ\).
Source: Myschool Past Questions Archive (Record #52594)