F3 Maths Problems

1.
(a)
AD // BC (given)
∠DBC = ∠ADB = 60° (alt. ∠s,AD // BC)

In ΔBDC :
sin∠DBC / CD = sin∠BDC/ BC (sine law)
sin60° / √6 = sin∠BDC / 2
sin∠BDC = sin60° × (2/√6)
sin∠BDC = (√3/2) × (2/√6)
sin∠BDC = 1/√2
∠BDC = 45°

(b)
∠BDC = ∠ABD = 45°
AB // DC (alt. ∠s equal)

Since AB // DC and AD // BC,
ABCD is a parallelogram (two pairs of opp. sides //)


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2.
(a)
In ΔBEC :
BC² = CE² +BE²
BC² = 2² + 4²
BC² = 20
BC = 2√5

In ΔCDE :
CD² = DE² +BE²
CD² = 1² + 2²
BC² = 5
BC = √5

(b)
tan∠CDE = CE/DE
tan∠CDE = 2/1
tan∠CDE = 2

tan∠BCE = 4/2
tan∠BCE = 2

tan∠CDE = tan∠BCE= 2
∠CDE = ∠BCE

In ΔBEC :
∠BCE + 90° + ∠EBC= 180° (∠ sum of Δ)
∠BCE + ∠EBC = 90°
But ∠ABE + ∠EBC = 90° (given)
Hence, ∠ABE = ∠BCE (axiom)

Since ∠CDE = ∠BCE and ∠ABE= ∠BCE,
then ∠ABE = ∠CDE
Hence, AB // DC (alt. ∠s equal)

(c)
AB // CD (proven)
∠EAB = ∠ECD (alt. ∠s,AB // CD)
∠ABE = ∠CDE (alt. ∠s,AB // CD)
ΔABE ~ ΔCDE (AAA)

AB / CD = BE / DE (corr. sides, ΔABE ~ ΔCDE)
AB / √5 = 4 / 1
AB = 4√5

Area of trapezium ABCD
= (CD + AB) × BC / 2
= (√5 + 4√5) × 2√5 / 2
= 25 square units

如果你唔知道咩係sin law，你或且可以試下以下呢1個方法

1a
Joint CH such that CH⊥BD
∠CBH=∠ADB=60° (alt. ∠s,AD // BC)

CH=BC sin ∠CBH
CH=2 sin 60°=√3

sin ∠CDH=CH/CD
sin ∠CDH=(√3)/(√6)=1/(√2)
∠CDH=45°