急! F5 Trigonometry 11q10

第一條:
26a.
OB=(64-x)cm
[因為佢話條string有64cm,姐係OAB=64 所以OB=64-x]

26b.
(sorry我唔記得F5學左cosine rule未,所以我呢度照用)
佢比左角OAB,OA,AB同OB. 所以by using cosine rule:
cos(60)=[x^2+(64-x)^2-40^2]/[2x(64-x)]
約完就得出:
x^2-64x-832=0
x=45.86 or x=18.14
因為題目話AO is shorter than OB
所以AO=18.14cm and OB=45.86cm (如果我無計錯)

第二條:
27a.
因為angle CPA=20, BPA=12 所以angle CPB=20-12=8
因為angle CPB= angle CQB
所以BCPQ are concyclic(converse of angles in the same segment)

27bi.
因為angle CAP=90
所以angle PAC=180-90-20=70
所以angle QCP=70-12=58
所以angle PQC=180-20-58=102
所以angle QBP=180-12-102-8=58
by using sine rule:
16/sin(58)=PB/sin(102+8)
PB=17.72904391m
17.72904391=AB/sin(12)
AB=3.69m

27bii.
PB=17.72904391m
所以
17.72904391/sin(58+12)=CB/sin(8)
CB=2.63m


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