中五數學求解答(20分)

在三角形ABC中，b:c=3:2，a=14cm及A=60度
求b,c

b : c = 3 : 2
設 b = 3k cm 及 c = 2k cm

根據餘弦定律：
a² = b² + c² - 2bc●cosA
14² = (3k)² + (2k)² - 2(3k)(2k)cos60°
9k² + 4k² - 6k² = 196
7k² = 196
k² = 28
k = √28 or k = -√28 (捨棄)
k = 2√7 cm

b = 3k cm = 6√7 cm
c = 2k cm = 4√7 cm


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[(5 - i)/(2 - i)]²
= [(5 - i)(2 + i)/(2 - i)(2 + i)]²
= [(10 + 5i - 2i - i²)/(4 - i²)]²
= [(11 + 3i)/5]²
= (11 + 3i)²/25
= (121 + 66i + 9i²)/25
= (112 + 66i)/25
= (112/25) + (66/25)i


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求 x

∠BAD = 90° (半圓內圓周角)
∠ABD + ∠ADB + ∠BAD = 180° (Δ內角和)
90° + ∠ADB + ∠BAD = 180°
∠ADB + ∠BAD = 90°
但 ∠ABD = ∠ADB (對等弧圓周角)
所以 ∠ADB = ∠BAD = 45°

∠CAD + 75° + ∠ADB = 180° (Δ內角和)
∠CAD + 75° + 45° = 180°
∠CAD = 60°

∠CAD/弧CD = ∠ADB/弧AB (圓周角與所對弧長成正比)
60°/(x cm) = 45°/(3 cm)
所以 x = 4


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在圖中,角PQS=30度,角QRS=75度及弧PQ=弧QR.求角RQS

∠PRS = ∠PQS = 30° (對同弧圓周角)

∠QRS = ∠QRP + ∠PRS
75° = ∠QRP + 30°
∠QRP = 45°

弧PQ = 弧QR (已知)
∠QSR = ∠QRP = 45° (對等弧圓周角)

∠RQS + ∠QRS + ∠QSR = 180° (Δ內角和)
∠RQS + 75° + 45° = 180°
∠RQS = 60°