(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°, < ABC = 120°) and (< CHG = 7x°), find the value of (< GHB).
Explanation
(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°) (angles on a straight line)
(< HBN = 180 – 7x°) (alternate angles)
(< EBN = 3x°) (corresponding angles)
Then (3x° + (180 – 7x)° = 120°)
(implies 3x° – 7x° + 180° = 120°)
(implies 180° – 120° = 7x° – 3x°)
(60° = 4x°)
(x = 15°)
Hence, (< GHB = 180° – (7 times 15°) = 180° – 105°)
= (75°).