附加數問題

(22 a) ∠CAD = ∠CAB - ∠DAB = 44°
∠OAD = (1/2)∠CAD (tangent properties)
= 22°
OD/OA = sin 22°
OA = r/sin 22°
∠FBE = ∠FBA - ∠EBA = 52°
∠OBE = (1/2)∠FBE (tangent properties)
= 26°
OE/OB = sin 26°
OB = r/sin 26°
(b) ∠OAB = ∠OAD + ∠DAB = 56°
∠OBA = ∠OBE + ∠EBA = 78°
∠AOB = 180° - ∠OAB + ∠OBA = 46°
Applying sine law in △OAB:
OB/sin ∠OAB = AB/sin ∠AOB
r/(sin 26° sin 56°) = 200/sin 46°
r = 101 m
(23) ∠C = 105° and ∠D = 60° for the reason of opp. angles of cyclic quad.
Join DB such that ∠CDB = ∠BDA = 30° (Equal chord BC = BA)
Also join OA, OB and OC and we have:
∠BOA = 2 × ∠BDA = 60° (Angle at centre = Twice the angle at circumference)
Therefore △OAB and △OBC are equilateral triangles as all angles are 60°.
So, AB = BC = r
Join AC and ∠BAC = 30° since △BAC is an isos. triangle.
Then ∠CAD = 45° and ∠COD = 90° (Angle at centre = Twice the angle at circumference)
Therefore CD = r√2 using the Pyth. theorem.
∠DOA = 360° - 90° - 60° - 60° = 150°
Using cosine law:
AD2 = 2r2 - 2r2 cos 150°
= (2 + √3)r2
AD = r√(2 + √3)

我未學，so我唔明

22.
(a)
angleCAD = 78 - 34 = 44 degrees
As AC and AD and tangents to the pond, OA bisects angleCAD
Thus, angleOAC = 44/2 = 22 degrees
OA = r/sin22
Similarly, OB = r/sin26
(b)
angleCAB + angleFBA = 78 + 102 = 180 degrees
So, AC // BF, hence, COF is a straight line
angleBOF = 90 - 26 = 64 degrees
angleAOC = 90 - 22 = 68 degrees
Thus, angleAOB = 180 - 64 - 68 = 48 degrees
By cosine law, AB^2 = OA^2 + OB^2 - 2(OA)(OB)cos(angleAOB)
120^2 = (r/sin22)^2 + (r/sin26)^2 - 2(r/sin22)(r/sin26)cos(48)
After solving the equation, r = 59 m (to the nearest meter)

23.
After finding the angles at the center, you can find the length of each side.
As, AB=BC, OB bisects angleABC
Hence, angleAOB = angleBOC = 60 degrees
(triangles AOB and BOC are equilateral triangles)
angleDAO = 75 - 60 = 15 degrees
angleADO = 15 degrees (bases angles of isos. triangle)
angleDOA = 180 - 15 - 15 = 150 degrees
angleCOD = 360 - 150 - 60 - 60 = 90 degrees

AB = BC = r
CD = sqrt(2) r
DA = sqrt[r^2 + r^2 - 2(r)(r)cos150] = r sqrt[2 +sqrt(3)] = r [sqrt(1/2) + sqrt(3/2)]

2008-01-17 23:34:05 補充：
(b)

∠OBA = ∠OBE + ∠EBA = 78° ..... should be 76°

∠AOB = 180° - ∠OAB + ∠OBA = 46° ......
should be ∠AOB = 180° - (∠OAB + ∠OBA) = 48°

仲有 ... AB = 120 ... not 200 ....

唉....後面明明有個o岩既 .... 居然選個錯既個最佳 .....
靚o的就最好 ?