0___0 中4maths MCCCCCCCCCC 題 6

46) ∠ACB = 90° (Angle in semi-circle)
x = 180° - 90° - 34° = 56°
Ans = C
47) ∠BAC = 25° (Angle at centre is twice of angle at circumference)
x + y + 25° = 50°
x + y = 25° (ans = A)
48) Join OB, then
∠AOB = 108° and ∠COB = 72° according to the given ratio.
Hence ∠OBE = 36° and ∠ODE = 36° (Base angles of isos. triangle)
∠BOD = 108° and therefore ∠COD = 36°
So arc DC:CB = 1:2 (ans = B)
49) Let ∠ACB = 2x, ∠ABC = 5x and ∠CAB = 3x
Then total = 180°
10x = 180°
x = 18°
∠BAC = 54°
50) w = ∠DAC, x = ∠ABD, y = ∠BCA and z = ∠BDC
For the reason of angle in the same segment,
∠BAC = z and ∠CBD = w
Therefore, w + x + y + z = 180° as they are angles in the same triangle.
51) ∠ACB = 25° (Angle at centre = Twice the angle at circumference)
∠BCO = 45° = ∠CBD (Base angles of isos. triangle)
Ans = A
53) ∠EBC = 60°
∠EBC = 120° (Opp angles of cyclic quad.)
Ans = B

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