+502 http://papers.xtremepapers.com/CIE/Cambridge%20IGCSE/Mathematics%20(0580)/0580_s12_qp_41.pdf
Question number 4
+129907 (a) Since AOC is a diameter, then angle ABC is an inscribed angle that measures (1/2) of is intercepted arc......and is intercepted arcc = 180°....so ABC = 90”
(b) find AC = sqrt [ 10 ^2 + 7^2 ] = sqrt [149]
So ..... 7/ sin ACB = sqrt [149] → sin ACB = 7 / sqrt [149]
→ arcsin (7 / sqrt [149] ) = ACB = 34.99° = 35°
(c) ADB = ACB because they are inscribed angles intercepting the same minor arc, AB
(d) AD / sin 77 = 7 / sin 35 → AD = 7 sin 77 / sin 35 ≈ 11.9
(e) Angle DAB = (180 - 77 - 35) = 68°
Area of ABD ≈ (1/2) (7)(11.9)sin(68) ≈ 38.6 cm^2
(f) ΔDEA ≈ ΔCEB
Area ΔAED = (1/2)(11.9)(DE)sin(35) = 12.3 → DE = 12.3 / [ (1/2)(11.9)sin(35) ] ≈ 3.6
So....AD/DE = BC/CE → 11.9 / 3.6 = 10/ CE → 3.6 / 11.9 = CE / 10 →
CE = 10 * 3.6 / 11.9 ≈ 3.03
So area of ΔBEC = (1/2)(3.03)(10) sin (35) ≈ 8.69 cm^2
