The three ratios
SOH CAH TOA
sin θ = oppositehypotenuse
cos θ = adjacenthypotenuse
tan θ = oppositeadjacent
30-60-90 and 45-45-90
Easy
In a 30-60-90 triangle, what is sin 30°?
sin 30° = opphyp = 12
Answer: 12
What is cos 30°?
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cos 30° = adjhyp = √32
Answer: √32
What is tan 45°?
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tan 45° = oppadj = 11
tan 45° = 1
Answer: 1
What is sin 60°?
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sin 60° = opphyp = √32
Answer: √32
What is cos 45°?
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cos 45° = adjhyp = 1√2
cos 45° = √22
Answer: √22
What is tan 30°?
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tan 30° = oppadj = 1√3
tan 30° = √33
Answer: √33
Medium
Given csc θ = √2 and 0 ≤ θ ≤ π2, find sin θ, cos θ, and tan θ.
csc θ = √21
csc θ is sin θ's friend, so we flip csc θ to find sin θ.
sin θ = 1√2
cos θ = adjhyp
cos θ = 1√2
cos θ = √22
tan θ = oppadj
tan θ = 1
Answer: sin √22, cos √22, tan 1
Given sec θ = 2 and 0 ≤ θ ≤ π2, find sin θ, cos θ, and tan θ.
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sec θ = 21
sec θ is cos θ's friend, so we flip the fraction.
cos θ = 12
sin θ = opphyp
sin θ = √32
tan θ = oppadj
tan θ = √31
tan θ = √3
Answer: sin √32, cos 12, tan √3
Given tan θ = 1 and 0 ≤ θ ≤ π2, find sin θ, cos θ, and tan θ.
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tan θ = 11
sin θ = opphyp
sin θ = 1√2
sin θ = √22
cos θ = adjhyp
cos θ = 1√2
cos θ = √22
tan θ = 1
Answer: sin √22, cos √22, tan 1
Given cos θ = −12 and 0 ≤ θ < π, find sin θ, cos θ, and tan θ.
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cos θ = −12
0 ≤ θ < π and cosine is negative, so Quadrant II (120°).
The reference triangle is 30-60-90 with adjacent 1, hypotenuse 2, opposite √3.
Sine is positive in Quadrant II.
sin θ = opphyp
sin θ = √32
Tangent is negative in Quadrant II.
tan θ = oppadj
tan θ = √3
tan θ = −√3
Answer: sin √32, cos −12, tan −√3
Given cot θ = √3 and 0 ≤ θ < π, find sin θ, cos θ, and tan θ.
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cot θ = √31
cot θ is tan θ's friend, so we flip the fraction.
tan θ = 1√3
sin θ = opphyp
sin θ = 12
cos θ = adjhyp
cos θ = √32
Answer: sin 12, cos √32, tan √33
Given sec θ = −√2 and 0 ≤ θ < π, find sin θ, cos θ, and tan θ.
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sec θ = −√21
sec θ is cos θ's friend, so we flip the fraction.
cos θ = 1−√2
The hypotenuse is always positive.
Sine is positive in Quadrant II.
sin θ = opphyp
sin θ = 1√2
sin θ = √22
Tangent is negative in Quadrant II.
tan θ = oppadj
tan θ = 1
tan θ = −1
Answer: sin √22, cos −√22, tan −1
Hard
A 30-60-90 triangle has hypotenuse 10. Find the side opposite 30°.
sin θ = opphyp
opphyp = 12 = x10
Solve for x.
x = 10 · 12
x = 5
Answer: 5
What is the side length opposite of 60° if the hypotenuse is 10?
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sin θ = opphyp
opphyp = √32 = x10
Solve for x.
x = 10 · √32
x = 5√3
Answer: 5√3
A 45-45-90 triangle has a leg of 7. Find the hypotenuse.
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sin θ = opphyp
opphyp = 1√2 = 7x
Solve for x.
x = 7 · √21
x = 7√2
Answer: 7√2
sin θ = 725. Find cos θ.
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sin θ = opphyp
opphyp = 725
opp = 7, hyp = 25
adj = √(25² − 7²)
adj = √(625 − 49)
adj = √576
adj = 24
cos θ = adjhyp
cos θ = 2425
Answer: 2425
cos θ = 2029. Find sin θ.
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cos θ = adjhyp
adjhyp = 2029
adj = 20, hyp = 29
opp = √(29² − 20²)
opp = √(841 − 400)
opp = √441
opp = 21
sin θ = opphyp
sin θ = 2129
Answer: 2129
A 30-60-90 triangle has the side opposite 30° equal to 4. Find the hypotenuse.
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sin θ = opphyp
opphyp = 12 = 4x
Solve for x.
x = 4 · 21
x = 8
Answer: 8
tan θ = 815. Find all six trig ratios.
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tan θ = oppadj
oppadj = 815
opp = 8, adj = 15
hyp = √(8² + 15²)
hyp = √(64 + 225)
hyp = √289
hyp = 17
sin θ = opphyp
sin θ = 817
cos θ = adjhyp
cos θ = 1517
tan θ = oppadj
tan θ = 815
csc θ = hypopp
csc θ = 178
sec θ = hypadj
sec θ = 1715
cot θ = adjopp
cot θ = 158
Answer: sin 817, cos 1517, tan 815, csc 178, sec 1715, cot 158
A right triangle has opposite 5 and hypotenuse 11. Find the missing side, then all six trig functions of θ.
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Find the missing leg using the Pythagorean theorem: a² + b² = c²
5² + b² = 11²
25 + b² = 121
b² = 121 − 25
b² = 96
b = √96
b = √(16 · 6)
b = 4√6
sin θ = opphyp
sin θ = 511
cos θ = adjhyp
cos θ = 4√611
tan θ = oppadj
tan θ = 54√6
tan θ = 5√624
csc θ = hypopp
csc θ = 115
sec θ = hypadj
sec θ = 114√6
sec θ = 11√624
cot θ = adjopp
cot θ = 4√65
Answer: 4√6; sin 511, cos 4√611, tan 5√624, csc 115, sec 11√624, cot 4√65
Given sin θ = 513 and cos θ > 0, find the other five trig functions.
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Sine is positive and cosine is positive, so Quadrant I.
sin θ = opphyp = 513
Find the missing leg using the Pythagorean theorem: a² + b² = c²
5² + b² = 13²
25 + b² = 169
b² = 169 − 25
b² = 144
b = √144
b = 12
cos θ = adjhyp
cos θ = 1213
tan θ = oppadj
tan θ = 512
csc θ = hypopp
csc θ = 135
sec θ = hypadj
sec θ = 1312
cot θ = adjopp
cot θ = 125
Answer: sin 513, cos 1213, tan 512, csc 135, sec 1312, cot 125
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