Maths problems [mixed 2]

(19) Let DX = 2m and CX = m, then
△XDY ~ △ABY for the reason of AAA and their side ratios are 2:3 with YX:AY = 2:3.
So let the height of △XDY and △ABY be 2h and 3h respectively, then area of ABCD = (3m)(5h) = 15mh
Now,
Area of △ADX = 5mh, so area of △DAY is 3/5 of it since AY:YX = 3:2.
Hence area of △DAY = 3mh
Also area of △XDY and △ABY are 2mh and 9mh/2 respectively.
So area of XYBC = 15mh - 3mh - 2mh - 9mh/2 = 11mh/2
Therefore area of △DAY:XYBC = 3:11/2 = 6:11
(21) Vertical height of A above the ground = 2x sin (90 - θ) = 2x cos θ
Vertical height of C above A = x sin θ
Thus ans = 2x cos θ + x sin θ
(22) From P and R, draw perp. lines to the building and let the feet of perp. formed then be C and D respectively.
From the given, we can let, and also have the followings:
AC = x, CD = h, DB = h, BQ = s
tan 25 = x/s
tan 50 = (x + 2h)/s = tan 25 + 2(h/s), giving h/s = (tan 50 - tan 25)/2
So the angle of elevation of A from R is ∠ARD where
tan ∠ARD = (x + h)/s
= tan 25 + (tan 50 - tan 25)/2
= (tan 50 + tan 25)/2
∠ARD = 39.7
(48) From O, drop perp. lines to AB and CD and let the feet then formed be E and F respectively.
Then E and F are mid-points of AB and CD resp.
So OE = OF = √(36 - 25) = √11 cm
Hence OEMF is a square with OM = √22 cm
(50) OD and OB are perp. to CE and AC respectively.
So ∠BCD = 360 - 90 - 90 - 85 = 95
Thus ∠AFE = 180 - 95 = 85 since ACEF is a cyclic quad.
(44) AB = BC tan 20 and BG = BC tan 65
By Pyth. thm:
AG = √(AB2 + BG2)
= BC √(tan2 20 + tan2 65)
= 12√(tan2 20 + tan2 65)
= 26 cm
(46) Let O be the centre, then after joining BO, we have ∠BOC = 80, i.e. arc BC subtends an angle of 80 at centre.
So arc ACB subtends an angle of 260 at centre and therefore its length is 6 (260/80) = 19.5

2009-06-27 22:56:15 補充：
(51) With ∠ODE = ∠OFE = 90, ODEF is a cyclic quad.
AD is perp. to CE since OD is perp. to CE
Also ∠CAD = ∠EAD for tangent properties.
∠ADC = ∠ADE = 90
So △ADC and △ADE are congurent (ASA)

2009-06-27 22:56:25 補充：
(51 cont'd)
So CD = DE and since BC = CD and EF = DE for tangent properties, BC = EF.
∠AOF = ∠AOB (Tangent properties)
So, ∠BOE = 2∠AOF
∠BOE = 2∠BDF (Angle at centre = Twice angle at circumference)
So ∠AOF = ∠BDF

2009-06-27 22:56:40 補充：
(52) ∠ABX = ∠DCX and ∠XAB = ∠XDC (Angles in the same segment)
So, △ABX ~ △DCX
∠ABX = ∠ACD (Angles in the same segment)
∠XAB = ∠DAC (Given)

2009-06-27 22:56:54 補充：
(52 cont'd)
So, △ABX ~ △ACD
∠ADX = ∠ACB (Angles in the same segment)
∠DAX = ∠CAB (Given)
So, △ADX ~ △ACB

2009-06-27 22:57:11 補充：
(54) Join the centres and the line's length will be r + 4
Also, the horizontal distance between the centres = 25 - r - 4 = 21 - r
And the vertical distance between the centres = r - 4
So by Pyth. thm:
(r - 4)^2 + (21 - r)^2 = (r + 4)^2

2009-06-27 22:57:29 補充：
(54 cont'd)
2r^2 - 50r + 457 = r^2 + 8r + 16
r^2 - 58r + 441 = 0
(r - 49)(r - 9) = 0
r = 49 (rejected) or 9

好詳細啊! thx so much!

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