Every right-angled trig question starts with naming the three sides relative to a chosen angle. Pick the acute angle θ in the question, then label:
GDC in DEGREE mode: AI SL uses degrees throughout. Switch to degree mode at the start and never change it. Radian-mode answers are silently wrong.
Worked examples
WE 1Pythagoras — ladder against a wall (find hypotenuse)
A ladder rests against a vertical wall. The base of the ladder is 1.4 m from the wall and the top reaches a point 4.8 m up the wall. Find the length of the ladder.
Sketch — right angle at the foot of the wall
a = 1.4 (base), b = 4.8 (height)
ladder = c = hypotenuse
Apply Pythagoras: c² = a² + b²
c² = 1.4² + 4.8²
= 1.96 + 23.04
= 25
c = √25 = 5 m
ladder = 5 m
(1.4, 4.8, 5) is the (7, 24, 25) triple multiplied by 0.2 — that’s why the answer is a clean integer. Spotting scaled triples saves time.
WE 2Pythagoras — guy wire on a pole (find a shorter side)
A vertical antenna pole is supported by a single guy wire of length 17 m, attached from the top of the pole to a point 8 m from the base of the pole on level ground. Find the height of the pole.
Sketch — right angle at the base of the pole
hypotenuse c = 17 (the wire)
base = 8, pole height = h (unknown)
Rearrange Pythagoras: h² = c² − a²
h² = 17² − 8² = 289 − 64
= 225
h = √225 = 15 m
pole height = 15 m
(8, 15, 17) is a Pythagorean triple. When you subtract two squares and get a square, you’ve probably spotted one.
WE 3SOH-CAH-TOA — find a side using sine
A wheelchair ramp is 8 m long and makes an angle of 30° with the horizontal ground. Find the vertical height the ramp reaches.
Sketch — angle 30° at the ground, hypotenuse = ramp
H = 8 (the ramp = hypotenuse)
O = h (vertical rise — opposite 30°)
Use SOH (have H, want O)
sin(30°) = O/H = h/8
h = 8 sin(30°)
= 8 × 0.5 = 4 m
vertical rise = 4 m
sin(30°) = 0.5 exactly — one of three special angles (30°, 45°, 60°) where the value is a clean number. Memorise these three.
WE 4SOH-CAH-TOA — find an angle
A child slides 10 m down a straight slide and lands at a point that is 6 m horizontally from the start of the slide. Find the angle the slide makes with the horizontal.
Sketch — right angle at the base, angle θ at the bottom of the slide
H = 10 (slide = hypotenuse)
A = 6 (horizontal distance — adjacent to θ)
Use CAH (have A and H, want θ)
cos θ = A/H = 6/10 = 0.6
θ = cos⁻¹(0.6)
≈ 53.13° (GDC, degree mode)
θ ≈ 53.1° (3 sf)
the slide is the 6-8-10 triangle (= 3-4-5 scaled by 2). The vertical drop, by Pythagoras, would be 8 m — useful check.
WE 53D — space diagonal and its angle with the base
A rectangular box has dimensions 6 m by 8 m on the floor and 5 m in height. Find (a) the length of the space diagonal (corner-to-opposite-corner), (b) the angle the space diagonal makes with the floor.
(a) Step 1 — find the floor diagonal (2D Pythagoras)
d_floor = √(6² + 8²) = √100 = 10 m
Step 2 — apply Pythagoras again with the height
d_space² = 10² + 5² = 100 + 25 = 125
d_space = √125 = 5√5 ≈ 11.2 m
(b) angle θ between diagonal and floor
opposite = 5 (vertical), adjacent = 10 (floor diag)
tan θ = 5/10 = 0.5
θ = tan⁻¹(0.5) ≈ 26.6°
(a) d = 5√5 ≈ 11.2 m · (b) θ ≈ 26.6°
3D Pythagoras = 2D Pythagoras twice. The “leg” of the second triangle is the floor diagonal you just found.
WE 6Pyramid — vertical height from slant edge
A square-based pyramid has a base of side 8 cm. The slant edge from the apex to a corner of the base is 9 cm. Find the vertical height of the pyramid.
Step 1 — find the half-diagonal of the base
full diagonal = 8√2 cm (Pyth: √(8²+8²) = √128)
half-diagonal = 4√2 cm
Step 2 — apply Pythagoras in the vertical triangle
slant edge² = h² + (half-diagonal)²
9² = h² + (4√2)²
81 = h² + 32
h² = 49
h = 7 cm
vertical height = 7 cm
the apex sits directly above the centre of the base. The “vertical triangle” goes from a base corner → centre of base → apex. Its legs are the half-diagonal and the vertical height; its hypotenuse is the slant edge.
💡 Top tips
- Always sketch and label H, O, A on the triangle relative to your chosen θ. This single step eliminates the most common mistakes.
- Pythagorean triples: (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25). Scaled versions count too — (6, 8, 10), (1.4, 4.8, 5), (9, 12, 15).
- Special angles: sin 30° = 0.5, sin 60° = √3/2, sin 45° = √2/2. Same values for cos with 30° and 60° swapped.
- 3D problems = 2D twice: find a 2D diagonal, then use it as the “leg” of a new 2D right triangle.
- For 3D pyramids: the “vertical triangle” links a base point → centre of base → apex. Use Pythagoras inside it.
⚠ Common mistakes
- Treating any side as the hypotenuse: only the side opposite the right angle is the hypotenuse. It’s always the longest.
- Forgetting to square-root: c² = 25 means c = 5, not 25. Pythagoras gives the square; finish the job.
- GDC in radian mode: AI SL works in degrees. sin(30°) = 0.5 in DEG mode; in RAD mode it returns sin(30 radians) ≈ −0.988 — totally wrong.
- Choosing the wrong ratio: write down the two side-letters you have / want, then match to SOH-CAH-TOA. Don’t guess.
- Confusing “slant edge” and “slant height” in a pyramid: slant edge goes corner-to-apex; slant height goes mid-base-edge-to-apex. They give different right triangles inside the pyramid.
Up next: Sine Rule, Cosine Rule & Area of a Triangle. SOH-CAH-TOA only works in right-angled triangles. For any other triangle, you’ll use the sine rule (opposite pairs), the cosine rule (two sides + included angle, or all three sides), and the area formula (1/2)ab sin C. A flowchart tells you which to pick.
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