Part 1 · Foundations
BasicsThe sine of angle θ — look at the triangle diagram. The red side is opposite θ, and the blue line is the hypotenuse. Which ratio is sin θ?
Memory key: SOH — Sine = Opposite / Hypotenuse
📐 Explanation
SOH = Sine → Opposite over Hypotenuse.
sin θ = Opposite / Hypotenuse
In the diagram: red side (Opposite) is across from θ. Blue side (Hypotenuse) is the longest, always opposite the right angle □.
For cosine, the diagram highlights the green side (adjacent — it touches θ) and the blue hypotenuse. Which ratio is cos θ?
Memory key: CAH — Cosine = Adjacent / Hypotenuse
📐 Explanation
CAH = Cosine → Adjacent over Hypotenuse.
cos θ = Adjacent / Hypotenuse
Key rule: Adjacent always touches the angle θ. Opposite is always across from θ. These two are the most commonly confused sides!
tan θ only needs the two legs — the hypotenuse is crossed out in the diagram! Which ratio is tan θ?
Memory key: TOA — Tangent = Opposite / Adjacent (no hyp!)
📐 Explanation
TOA = Tangent → Opposite over Adjacent. No hypotenuse!
tan θ = Opposite / Adjacent
Also works as: tan θ = sin θ ÷ cos θ. Tangent can be greater than 1 — unlike sin and cos which are always ≤ 1.
In right △ABC with right angle at C, which side is the hypotenuse? Look at the diagram — the right angle □ is at C.
Memory key: Hyp = always opposite the right angle □
📐 Explanation
The hypotenuse is always opposite the right angle. Right angle = C → side opposite C = AB.
AB is the hypotenuse
Common trap: picking the "longest-looking" side visually. Always locate the right angle box □ first, then the hypotenuse is the side directly across from it.
Find sin 30° using the 30-60-90 triangle. The sides are in ratio 1 : √3 : 2. The side opposite 30° is labeled in red.
Memory key: 30-60-90 sides: 1, √3, 2(hyp). sin 30° = opposite/hyp.
📐 Explanation
Opposite to 30° = 1, Hypotenuse = 2.
sin 30° = 1 / 2 = 0.5
Trap: √3/2 = sin 60°. √2/2 = sin 45°. These three special angles appear in nearly every test — memorize the full table: sin 30°=½, sin 45°=√2/2, sin 60°=√3/2.
Part 2 · Core Practice
Frequently MissedA right triangle: legs 3 and 4, hyp 5. What is sin θ for the angle θ that is opposite the side of length 3?
Memory key: 3-4-5 triple — most common triangle! θ is at bottom-right here.
📐 Explanation
θ is opposite side 3 → Opp = 3, Hyp = 5.
sin θ = 3/5 = 0.6
Option B (4/5) is cos θ — that uses the adjacent side (4). Always ask: "which side is directly across from my angle?"
Same 3-4-5 triangle. Now find tan θ for the angle opposite the side of length 4. The hypotenuse is greyed out — tan ignores it!
Memory key: TOA — Opp = 4, Adj = 3. Hyp is irrelevant for tangent!
📐 Explanation
Opp = 4, Adj = 3. Hypotenuse (5) is not used in tangent!
tan θ = 4/3 ≈ 1.33
Important: tan CAN be greater than 1. But sin and cos are always between 0 and 1 for acute angles.
If sin θ = 0.6, what is cos θ? (θ is acute)
The diagram shows WHY subtracting gives the wrong answer!
The diagram shows WHY subtracting gives the wrong answer!
Memory key: sin²θ + cos²θ = 1 — square them, don't subtract!
📐 Explanation
sin²θ + cos²θ = 1 → (0.6)² + cos²θ = 1 → 0.36 + cos²θ = 1 → cos²θ = 0.64
cos θ = √0.64 = 0.8
THE biggest trap in Grade 8 trig: "cos = 1 − sin = 0.4" is WRONG! You must square sin first, then subtract, then square root.
Find tan 45°. In a 45-45-90 triangle, both legs are equal. What happens when you divide a number by itself?
Memory key: 45-45-90 — equal legs! Opp = Adj → tan = 1.
📐 Explanation
In 45-45-90: both legs equal a. Opp = a, Adj = a.
tan 45° = a / a = 1
At 45°, sin = cos = √2/2 ≈ 0.707, and tan = 1. This is the only acute angle where sin equals cos!
Which is always true? The diagram shows the Pythagorean identity comes directly from Pythagoras' theorem — divide everything by hyp²!
Memory key: sin²+cos²=1 — squares, not the values themselves!
📐 Explanation
Pythagoras: opp² + adj² = hyp². Divide all by hyp² → sin²θ + cos²θ = 1.
sin²θ + cos²θ = 1 (always!)
Check option A: sin 30° + cos 30° = 0.5 + 0.866 = 1.366 ≠ 1. It's wrong for almost every angle!
Part 3 · Applied Problems
Real WorldA 10 m ladder leans against a wall at 60° to the ground. How high up the wall? The ladder is the hypotenuse, and the height is opposite to 60°.
Memory key: SOH — height = opposite, ladder = hyp. Use sin 60° = √3/2.
📐 Explanation
Height = opposite to 60°, Ladder = hyp (10 m). Use SOH:
height = 10 × sin 60° = 10 × (√3/2) = 5√3 ≈ 8.66 m
Options A & B both equal 5 m — they used the wrong angle (30°) or wrong function (cos). Always check which angle you're computing for!
You stand 50 m from a tree. Angle of elevation to the top = 30°. How tall is the tree? The 50 m is the horizontal distance — not the hypotenuse!
Memory key: TOA — you know adj (50m), want opp (height). Use tan!
📐 Explanation
adj = 50 m, angle = 30°, want height (opp). Use TOA:
tan 30° = height/50 → height = 50 × (1/√3) = 50/√3 ≈ 28.9 m
Option C (86.6 m) uses tan 60° — reading the wrong angle. Always label your angle clearly in the diagram!
A ramp rises 2 m over 8 m horizontal. Find angle θ. The diagram shows rise = opposite, run = adjacent. Use inverse tan to find the angle!
Memory key: arctan = tan⁻¹ — "undo" tangent to get the angle.
📐 Explanation
Opp = 2 (rise), Adj = 8 (run) → tan θ = 2/8 = 0.25.
θ = tan⁻¹(0.25) ≈ 14°
Option A flips the fraction: tan⁻¹(8/2) gives the complementary angle (76°). Option C: sin⁻¹ needs hypotenuse = √68, not the run (8).
A ship sails 40 km at 25° above east. How far north? The diagram shows the north component is the vertical (opposite) leg. sin 25° ≈ 0.423.
Memory key: North = Opposite to the east angle → SOH. Distance sailed = hyp!
📐 Explanation
North = opposite side, voyage = hyp (40 km). sin 25° = North / 40.
North = 40 × 0.423 = 16.92 km
Option A (36.24 km) is the east component (cos 25°). The diagram makes it clear: north = short vertical leg, east = long horizontal leg.
If cos θ = 5/13, find sin θ. (θ is acute)
The diagram shows: cos gives us adj = 5, hyp = 13. Find the missing leg!
The diagram shows: cos gives us adj = 5, hyp = 13. Find the missing leg!
Memory key: 5-12-13 triple! cos = adj/hyp, so adj=5, hyp=13 → opp = 12.
📐 Explanation
cos θ = 5/13 → adj = 5, hyp = 13. By Pythagoras: opp = √(13² − 5²) = √(169−25) = √144 = 12.
sin θ = 12/13
Memorize: 3-4-5 and 5-12-13 triples. These two triangles appear in over half of all trig test problems!
Part 4 · Challenge Level
Hard / TrickyTwo buildings are 30 m apart. From the top of the shorter one, the angle of elevation to the taller top is 40°. Find the height difference. (tan 40° ≈ 0.839)
Memory key: elevation = look UP. adj = 30m gap, want opp = height diff.
📐 Explanation
adj = 30 m (gap), angle = 40°, want opp (height diff). Use TOA:
diff = 30 × tan 40° ≈ 30 × 0.839 = 25.17 m
Option A (10.92) = 30 × tan 20° — that's the height of the short building, not the difference. Read carefully!
Angles A and B are complementary (A + B = 90°). The diagram shows both in the same triangle — notice what changes and what stays the same!
Memory key: CO-sine = complement's sine. sin A = cos(90°−A).
📐 Explanation
B = 90° − A → cos B = cos(90° − A) = sin A.
sin A = cos B (always when A + B = 90°)
Example: sin 30° = cos 60° = 0.5. The word "cosine" literally means "complement's sine" — this identity is built into the name!
A 120 m kite string: first at 55°, then pushed to 70°. By how much did the kite rise? Calculate both heights from the diagram, then subtract!
Memory key: height difference = h₂ − h₁ — calculate each SOH height, then subtract.
📐 Explanation
h₁ = 120 × sin 55° = 120 × 0.819 = 98.28 m (green kite)h₂ = 120 × sin 70° = 120 × 0.940 = 112.8 m (red kite) Rise = 112.8 − 98.28 = 14.52 m Options C & D are individual heights, not the difference! "By how much did it rise?" means you need subtraction.
Right △PQR, right angle at Q, angle P = 38°, hypotenuse PR = 25 cm. Find QR. The diagram labels all vertices — is QR opposite or adjacent to angle P?
Memory key: Draw, label, then SOH/CAH/TOA. QR is opposite to P → use sin!
📐 Explanation
QR is opposite angle P (38°). PR is hyp (25 cm). Use SOH:
QR = 25 × sin 38° ≈ 25 × 0.616 = 15.4 cm
Option A (19.7 cm) = 25 × cos 38° — that's the adjacent side RQ. The diagram makes this mix-up immediately visible!
A 6 m pole casts a 6√3 m shadow. What angle does the sun's ray make with the ground? The pole is vertical (opposite), shadow is horizontal (adjacent).
Memory key: tan = pole/shadow = 6/6√3 = 1/√3. Which special angle has tan = 1/√3?
📐 Explanation
Opp = 6 m (pole), Adj = 6√3 m (shadow). TOA:
tan θ = 6/6√3 = 1/√3 → θ = tan⁻¹(1/√3) = 30°
Master table: tan 30°=1/√3 ≈ 0.577, tan 45°=1, tan 60°=√3 ≈ 1.732. Shadow longer than pole → angle less than 45°. ✓
Quiz Complete! 🎉
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