F3 Maths Problems

[匿名]

2015-05-10 7:17 pm

1) In the figure, ABC is horizontal and V is vertically above A. The angle of elvation of V from B and C are 45° and 60° respectively. ∠BAC=90° and VA=10.
(a) Show that ∠ACB=60° .
(b) Find, correct to the nearest degree, the degree, the angle between the plane VBC and the planes ABC.
(c) D is point on BC such that the angle of elevation of V from D is 60° and CD>0. Find CD.
Figure:http://postimg.org/image/kmizjl02t/

2)In the fiugre, ABCD is a trapezium where ∠ADC-∠BCD=90°. E is a point on DC such that AB=DE=EC=1. BC=√3.
(a) Show that the area of triangle AEB is (√(3)+1)/2.
(b) Find ∠DEA and ∠BEC. Hence, find ∠AEB.
(c) Using above results, show that sin 75° =(√(6)+√(2))/4.
Figure:http://postimg.org/image/3qf1o423x/

Need Steps, plz!

回答 (1)

50418129

2015-05-12 12:19 am

✔ 最佳答案

(1)(a)

VA / AB = tan 45 => AB = 10
VA / AC = tan 60 => AC = 10/√3
tan∠ACB = AB / AC = √3 => ∠ACB = 60

(1)(b)

P is a point on BC such that AP perpendicular to BC
sin∠ACB = AP / AC => AP = 5
tan∠VPA = VA / AP => ∠VPA = 63.4

(1)(c)

cos∠ACP = CP / AC => CP = 5/√3
CD = 2CP = 10/√3

(2)(a)

area = (AD + BC) x CD / 2 - AD x DE / 2 - BC x CE / 2
= (1 + √3) - 1 / 2 - √3 / 2
= (1 + √3) / 2

2015-05-12 08:35:13 補充：
(2)(b)

tan∠DEA = AD / AE => ∠DEA = 45
tan∠BEC = BC / BE => ∠BEC = 60
∠AEB = 180 - 45 - 60 = 75

2015-05-12 08:35:47 補充：
(2)(c)

P is a point on BE such that AP perpendicular to BE
cos ∠BEC = EC / BE => BE = 2
cos ∠AED = DE / AE => AE = √2
sin ∠AEB = AP / AE => AP = √2 sin 75

area of triangle ABE = AP x BE / 2 = (1 + √3) / 2
√2 sin 75 x 2 = 1 + √3
sin 75 = (1 + √3) / 2√2 = (√2 + √6) / 4

參考: knowledge

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