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http://papers.xtremepapers.com/CIE/Cambridge%20IGCSE/Mathematics%20(0580)/0580_s12_qp_41.pdf

Question number 4

 Jan 21, 2017
 #1
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(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

 

 

cool cool cool

 Jan 21, 2017

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