(a) If (3 – x), 6, (7 – 5x) are consecutive terms of a geometric progression (GP) with constant ratio r > 0, find the :
(i) values of x ; (ii) constant ratio.
(b) In the diagram, |AB| = 3 cm, |BC| = 4 cm, |CD| = 6 cm and |DA| = 7 cm. Calculate <ADC, correct to the nearest degree.
Explanation
(a) (i) (3 – x), 6, (7 – 5x).
First term = a = (3 – x)
common ratio = (frac{6}{3 – x} = frac{7 – 5x}{6})
(36 = (3 – x)(7 – 5x))
(36 = 21 – 15x – 7x + 5x^{2})
(36 = 21 – 22x + 5x^{2})
(5x^{2} – 22x + 21 – 36 = 0)
(5x^{2} – 22x – 15 = 0)
(5x^{2} – 25x + 3x – 15 = 0 implies 5x(x – 5) + 3(x – 5) = 0)
((x – 5)(5x + 3) = 0)
(x = -frac{3}{5} text{ or = } 5)
(ii) Constant ratio = (frac{6}{3 – x} = frac{7 – 5x}{6})
Using x = 5,
(r = frac{7 – 5(5)}{6} = frac{-18}{6})
= (-3)
(b) Considering (Delta) AOC
(b^{2} = 4^{2} + 3^{2})
(b^{2} = 16 + 9 = 25)
(b = sqrt{25} = 5 cm)
Considering (Delta) ACD,
Using cosine rule,
(cos D = frac{a^{2} + b^{2} – c^{2}}{2ab} = frac{a^{2} + c^{2} – b^{2}}{2ac})
= (frac{6^{2} + 7^{2} – 5^{2}}{2 (6)(7)})
= (frac{36 + 49 – 25}{84})
(cos D = frac{60}{84})
(cos D = 0.714)
(D = cos^{-1} (0.714))
(D = 44.4° approxeq 44°) (to the nearest degree)