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Jika vektor oa = (1, 2) vektor ob = (4, 2) dan sudut teta adalah sudut antara vektor oa dan vektor ob, tan teta =

Jika vektor oa = (1, 2) vektor ob = (4, 2) dan sudut teta adalah sudut antara vektor oa dan vektor ob, tan teta =

Jawaban: tan θ = 3 /4

Ingat konsep berikut ini:
cos θ = (a⋅b)/|a||b|
vektor a = (x, y)
vektor b = (p, q)
a⋅b = (x, y) ⋅ (p, q) = px + qy
|a| = √(x²+y²)
sisi miring² = sisi samping² + sisi depan²
pada kuadran I tan bernilai positif

vektor oa = (1, 2) vektor ob = (4, 2) berada di kuadran I
vektor oa = (1,2)
|a| = √(1²+2²)
|a| = √(1+4)
|a| = √5

vektor ob = (4, 2)
|b| = √(4²+2²)
|b| = √(16+4)
|b| = √20
|b| = √(4 x 5)
|b| = 2√5

|a| ⋅ |b| = √5 ⋅ 2√5 = 10

oa ⋅ ob = (1,2) ⋅ (4,2)
oa ⋅ ob = (1⋅4) + (2⋅2)
oa ⋅ ob = 4 + 4
oa ⋅ ob = 8

cos θ = (a⋅b)/|a||b|
cos θ = 8/10
cos θ = 4/5
sisi samping = 4
sisi miring = 5
sisi miring² = sisi samping² + sisi depan²
sisi depan² = sisi miring² – sisi samping²
sisi depan² = 5² – 4²
sisi depan² = 25 – 16
sisi depan² = 9
sisi depan = ±√(9)
sisi depan = ±3
tan θ = 3 /4, atau
tan θ = – 3 /4
karena vektor oa dan ob ada pada kuadran I maka tan dipilih yang bernilai positif yaitu 3/4.

Jadi, tan θ = 3 /4.

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