△ Trigonometry — SOHCAHTOA

Trigonometry lets you find unknown sides and angles in right-angled triangles. Engineers use it constantly — for ramp angles, roof pitches, cable lengths, force components, and machine geometry.

Right-angled triangles only SOH CAH TOA Finding sides Finding angles (inverse trig) Angles of elevation & depression

The three sidesAlways named relative to YOUR chosen angle θ

Hypotenuse The longest side. Always opposite the right angle. Never changes regardless of which angle you call θ.

Opposite The side directly across from angle θ. Changes if you change which angle is θ.

Adjacent The side next to angle θ (not the hypotenuse). Changes if you change which angle is θ.

⚠️ The sides are NOT fixed names — Opposite and Adjacent swap when you move θ to a different corner!

Why engineers need trig

Ramp & roof angles — given horizontal distance and height, find the slope angle
Cable lengths — given height of a mast and angle, find the wire length
Force components — split a force along two axes using its angle
Machine geometry — find travel distances on angled surfaces
Setting out — calculate distances on site from angles and known lengths

📌 Trig only works on right-angled triangles. If your triangle has no right angle, you need the Sine Rule or Cosine Rule (covered in the next module).

📐

SOHCAHTOA — The Three Ratios

SOH CAH TOA is a memory aid for the three trigonometric ratios. Each one connects an angle to two sides of the triangle.

Memory tricks:
“Silly Old Harry — Caught A Herring — Trawling Off America”
Or just: SOHCAHTOA said as one word: “sock-ah-toe-ah”

Which ratio to use?The 3-question decision

Label the three sides relative to your angle θ: Hyp, Opp, Adj
Identify which two sides are involved (the one you know and the one you want)
Pick the ratio that uses exactly those two sides:
• Opp & Hyp → SOH (sine)
• Adj & Hyp → CAH (cosine)
• Opp & Adj → TOA (tangent)

💡 Pro tip: Write all three sides on the diagram first (H, O, A). Then tick which two are relevant. The ratio uses those two letters.

Quick reference table

OPP & HYP → sin θ = O/H

ADJ & HYP → cos θ = A/H

OPP & ADJ → tan θ = O/A

📏

Finding an Unknown Side

You know one side and one angle (θ) — and you want to find another side. Pick the right ratio, substitute the values, then rearrange to find the unknown side.

Method: ① Label H, O, A on the diagram ② Pick SOH/CAH/TOA ③ Write the equation ④ Rearrange ⑤ Calculate

Rearranging for the unknown sideThe unknown can be top or bottom of the fraction

There are two cases for finding a side:

Unknown on TOP: sin θ = x/Hyp → x = Hyp × sin θ

Unknown on BOTTOM: sin θ = Opp/x → x = Opp / sin θ

💡 Unknown on top? Multiply. Unknown on bottom? Divide.
This covers all six possible rearrangements!

Three worked examples

Find the hypotenuse: θ=40°, Opp=15m

Sides: OPP=15, HYP=? → SOH

Write: sin(40°) = 15 / H

Rearrange: H = 15 / sin(40°)

Calculate: H = 15 / 0.6428

Answer: H = 23.3 m ✅

Find the adjacent: θ=55°, Hyp=20m

Sides: ADJ=?, HYP=20 → CAH

Write: cos(55°) = A / 20

Rearrange: A = 20 × cos(55°)

Calculate: A = 20 × 0.5736

Answer: A = 11.5 m ✅

🧭

Finding an Unknown Angle

You know two sides and want to find the angle. The method is identical to finding a side — except the final step uses the inverse trig button on your calculator: sin⁻¹, cos⁻¹, or tan⁻¹.

Key step: Once you have sinθ = (a number), press SHIFT / 2nd → SIN−¹ on your calculator to find θ.

The inverse trig keysOn your calculator — SHIFT or 2nd then the trig button

sin θ = value → θ = sin⁻¹(value)

cos θ = value → θ = cos⁻¹(value)

tan θ = value → θ = tan⁻¹(value)

💡 On a Casio calculator: press SHIFT then sin / cos / tan. Make sure your calculator is in DEGREE mode (not radians!) — check for a D or DEG symbol on the display.

⚠️ If your calculator is in radian mode, sin⁻¹(0.8) = 0.927 instead of 53.1 — always check the mode first!

Three worked examples

Find θ: Adj = 12, Hyp = 15

Sides: ADJ=12, HYP=15 → CAH

Write: cos θ = 12/15 = 0.8

Inverse: θ = cos⁻¹(0.8)

Answer: θ = 36.9° ✅

Find θ: Opp = 7, Adj = 5

Sides: OPP=7, ADJ=5 → TOA

Write: tan θ = 7/5 = 1.4

Inverse: θ = tan⁻¹(1.4)

Answer: θ = 54.5° ✅

🏗

Word Problems — Construction & Access

Every word problem becomes easy with the same three steps: draw the triangle, label H/O/A, pick SOHCAHTOA. Always draw first!

🏗️ Problem 1 — Access Ramp

A wheelchair ramp must rise 1.2 m over a horizontal distance of 7.2 m.
Find the angle of the ramp to the horizontal and the length of the ramp surface.

Part a) Find the angle

Known: OPP = 1.2 m, ADJ = 7.2 m

Ratio: OPP & ADJ → TOA: tan θ = OPP / ADJ

Substitute: tan θ = 1.2 / 7.2 = 0.1667

Inverse tan: θ = tan⁻¹(0.1667)

Answer: θ = 9.5° ✅

Part b) Find the ramp length

Known: OPP = 1.2 m, ADJ = 7.2 m, HYP = ?

Use Pythagoras: H² = 1.2² + 7.2² = 1.44 + 51.84 = 53.28

Or trig: sin(9.5°) = 1.2 / H → H = 1.2 / sin(9.5°) = 1.2 / 0.1650

Answer: H = 7.3 m ✅

🏠 Problem 2 — Roof Pitch

A roof has a horizontal span of 8 m (half-span = 4 m) and a ridge height of 3 m above the wall plate.
Find the pitch angle and the rafter length.

Find pitch angle and rafter length

Known: OPP = 3 m (rise), ADJ = 4 m (half-span)

Angle (TOA): tan θ = 3/4 = 0.75

Inverse tan: θ = tan⁻¹(0.75)

Pitch angle: θ = 36.9° ✅

Rafter (SOH): sin(36.9°) = 3/H → H = 3/sin(36.9°) = 3/0.6004

Rafter length: H = 5.0 m ✅

📌 Check: 3² + 4² = 9 + 16 = 25 = 5² ✔ This is a 3-4-5 Pythagorean triple — exact answers!

🪜 Problem 3 — Ladder Safety

Safe ladder use requires an angle of 75° to the horizontal.
A ladder is 6 m long.
How high up the wall does it reach, and how far from the wall is the base?

Both parts — HYP=6m, angle=75°

Height (OPP): OPP & HYP → SOH: sin(75°) = H/6

Rearrange: H = 6 × sin(75°) = 6 × 0.9659

Height: H = 5.80 m ✅

Base (ADJ): ADJ & HYP → CAH: cos(75°) = A/6

Rearrange: A = 6 × cos(75°) = 6 × 0.2588

Base distance: A = 1.55 m from wall ✅

⚙

Word Problems — Engineering Contexts

These are the types you will see in T Level / Level 3 Engineering exams — forces, machine geometry, pipe runs, cable installations.

⚡ Problem 1 — Force Components

A force of 500 N acts at 35° above the horizontal.
Find the horizontal component (Fₕ) and vertical component (Fᵥ) of the force.

Find Fₕ and Fᵥ

Horizontal: ADJ & HYP → CAH: cos(35°) = Fₕ / 500

Rearrange: Fₕ = 500 × cos(35°) = 500 × 0.8192

Fₕ: Fₕ = 409.6 N ✅

Vertical: OPP & HYP → SOH: sin(35°) = Fᵥ / 500

Rearrange: Fᵥ = 500 × sin(35°) = 500 × 0.5736

Fᵥ: Fᵥ = 286.8 N ✅

✔ Check: Fₕ² + Fᵥ² = 409.6² + 286.8² = 167,773 + 82,254 = 250,027 ≈ 500² = 250,000 ✔

🛠️ Problem 2 — Angled Pipe Run

A pipe runs at 22° to the horizontal for a true length of 4.5 m.
Find the vertical rise and horizontal run of the pipe.

Find vertical rise and horizontal run

Known: HYP = 4.5 m, angle = 22°

Vertical rise: SOH: sin(22°) = rise / 4.5

Rise: rise = 4.5 × sin(22°) = 4.5 × 0.3746 = 1.69 m ✅

Horiz. run: CAH: cos(22°) = run / 4.5

Run: run = 4.5 × cos(22°) = 4.5 × 0.9272 = 4.17 m ✅

🔨 Problem 3 — Cutting Angle on a CNC Machine

A CNC milling cutter moves 85 mm horizontally and 32 mm vertically in one pass.
At what angle to the horizontal is the tool path?

Find the tool path angle

Known: ADJ = 85 mm, OPP = 32 mm

Ratio: OPP & ADJ → TOA: tan θ = 32/85

Calculate: tan θ = 0.3765

Inverse tan: θ = tan⁻¹(0.3765)

Answer: θ = 20.6° ✅

👁

Angles of Elevation & Depression

Two key types of angle used in surveying, construction and engineering — both measured from the horizontal.

Key facts

Angle of elevation = angle measured UPWARD from horizontal

Angle of depression = angle measured DOWNWARD from horizontal

💡 The angle of elevation from A to B equals the angle of depression from B to A — they are alternate angles!

📌 Both are always measured from the horizontal — never from the vertical.

Worked word problems

🏗 Crane height

From 60 m away, the angle of elevation to the top of a crane is 52°. Find the crane height.

Setup: ADJ=60m, angle=52°, OPP=height

TOA: tan(52°) = h/60

Height: h = 60×tan(52°) = 60×1.2799

Answer: h = 76.8 m ✅

📍 Distance from cliff

From the top of a 40 m cliff, the angle of depression to a boat is 18°. Find the distance from cliff base to boat.

Setup: OPP=40m, angle=18°, ADJ=distance

TOA: tan(18°) = 40/d

Rearrange: d = 40/tan(18°) = 40/0.3249

Answer: d = 123.1 m ✅

⚙️

Applications

Where this topic is used in engineering, manufacturing, maintenance and daily life.

🏗️Construction: Roof pitch height=span×tan(angle). Ladder angle 75°.
⚡Electrical: Phasor: V_R=Vcosφ, V_L=Vsinφ in AC circuit.
🔧Mechanical: Force on inclined plane: Fx=Fcosθ, Fy=Fsinθ.
🛩️Daily Life: Bearing navigation, ramp angles, satellite alignment.

🔨

Trigonometry in Engineering

Every branch of T Level / Level 3 Engineering uses trig. Here are the six most-tested applications.

⚖️ Resolving Forces

Any force at an angle must be split into horizontal and vertical components before applying equilibrium equations.

Fₕ = F × cos(θ) ← horizontal
Fᵥ = F × sin(θ) ← vertical

Example: F=200N at 40°
Fₕ = 200 × cos40° = 153.2 N
Fᵥ = 200 × sin40° = 128.6 N

🛠️ Cutting Tool Angles

CNC machines and manual lathes require precise tool angles. Setting the compound slide angle uses trig.

Taper angle: tan(θ) = (D−d) / (2L)

D=50mm, d=30mm, L=80mm:
tan θ = (50−30)/(2×80) = 20/160 = 0.125
θ = tan⁻¹(0.125) = 7.1°

🧱 Roof & Structural Geometry

Rafter lengths, hip angles, purlin positions — all calculated using SOHCAHTOA from known span and pitch.

Pitch angle: tan θ = rise/half-span
Rafter: L = rise / sin θ

Rise=2.5m, half-span=5m:
θ = tan⁻¹(0.5) = 26.6°
Rafter = 2.5/sin(26.6°) = 5.59 m

📡 Signal & Survey Angles

Survey instruments and antennas are positioned using elevation angles. The trig gives heights and distances.

Height = dist × tan(elevation angle)
Distance = height / tan(elevation angle)

Antenna at 25m, elevation 32°:
Horiz. dist = 25/tan(32°) = 40.0 m

🔧 Bolt Circle — PCD Layout

Holes on a Pitch Circle Diameter (PCD) are equally spaced. Finding the chord length between adjacent holes uses SOHCAHTOA.

n holes on PCD diameter D:
Half-angle = 360°/(2n) = 180°/n
Chord = D × sin(180°/n)

Example: 6 holes on PCD 120mm:
Chord = 120 × sin(30°)
= 120 × 0.5 = 60.0 mm

⚡ Electrical — Impedance Triangle

In AC circuits, resistance R, reactance X and impedance Z form a right-angled triangle. SOHCAHTOA gives the phase angle.

Z = √(R² + X²) (Pythagoras)
Phase angle φ = tan⁻¹(X/R)
Power factor = cos φ

Example: R=60Ω, X=80Ω:
Z = √(3600+6400) = 100 Ω
φ = tan⁻¹(80/60) = 53.1°
Power factor = cos53.1° = 0.6

🏎️ V-Belt Drive — Wrap Angle

The wrap angle of a V-belt on a pulley determines how much grip is available. It depends on the centre distance and pulley radii.

sin α = (R − r) / C
where R=large radius, r=small, C=centre distance

Wrap angle on small pulley:
β = 180° − 2α

Example: R=150mm, r=60mm, C=400mm:
sinα=(150−60)/400=0.225 → α=13.0°
β = 180°−26° = 154° wrap

📏 Dovetail Slide — Measuring Over Rollers

A dovetail slide has a 60° included angle. Inspection uses rollers of known diameter sitting in the groove. SOHCAHTOA gives the measurement across the rollers.

Half-angle of dovetail = 30°
Roller diameter = d, groove depth = h

Distance from centre of roller to groove wall:
x = (d/2) / tan(30°) = (d/2) × √3

Example: d=10mm:
x = 5 × 1.732 = 8.66mm
M = groove_width + d + 2×(d/2)/tan30°
Check with formula sheet in exam

📐

SOHCAHTOA Interactive Solver

Enter any two known values from a right-angled triangle and the solver finds the rest — with every step shown clearly.

📏 Find a Missing Side or AngleStep-by-step SOHCAHTOA

ℹ️ How to use: Use the dropdown to choose what you want to find. Type the two known values into the boxes. The full SOHCAHTOA working appears instantly below.

I want to find

Select what you want to find and enter the known values above.

👆 Try: Find O: angle=35°, H=10 · Find H: O=3, A=4 · Find angle: O=5, H=8

△ Triangle VisualiserRight-angled triangle

ℹ️ How to use: Drag the angle slider to change the angle θ. The triangle redraws live showing the ratio of Opposite, Adjacent and Hypotenuse — watch how the sides change as the angle grows.

Angle θ: 35°

      sin(35°) = 0.574  |  cos(35°) = 0.819  |  tan(35°) = 0.700
    

SOHCAHTOA at a glance

SOH sin(θ) = Opposite / Hypotenuse

CAH cos(θ) = Adjacent / Hypotenuse

TOA tan(θ) = Opposite / Adjacent

💡 To find the angle: use sin⁻¹, cos⁻¹ or tan⁻¹ on your calculator.

Choosing the right ratio

Know H, want O? → SOH: O = H sin(θ)

Know H, want A? → CAH: A = H cos(θ)

Know A, want O? → TOA: O = A tan(θ)

Know O & H? → angle = sin⁻¹(O/H)

Know A & H? → angle = cos⁻¹(A/H)

Know O & A? → angle = tan⁻¹(O/A)

🎮

Quick Fire Quiz

Test your knowledge — 10 questions, instant feedback.

⚡ Trigonometry Blitz

ℹ️ How to play: A trigonometry question appears. Click the correct value from the four options. Your score and streak update after each answer — click Next to continue.

Score: 0 Streak: 0 🔥 Q 1/10

Loading...

✏

Practice Questions

15 questions. Always: draw the triangle first, label H/O/A, write the ratio, show every step.

🟢 Tier 1 — Foundations

ℹ️ How to use: Work through each question and write down your answer. When ready, click Show Answer to reveal the full worked solution and mark scheme.

Q1

In a right-angled triangle, θ = 40° and the hypotenuse = 12 cm. Find the opposite side.

[2]

SOH: sin θ = Opp/Hyp → Opp = Hyp × sin θ

SOH: sin(40°) = Opp/12
Opp = 12 × sin(40°) = 12 × 0.6428
Opp = 7.71 cm

Q2

In a right-angled triangle, θ = 28° and the adjacent = 9 m. Find the hypotenuse.

[2]

CAH: cos θ = Adj/Hyp → Hyp = Adj / cos θ

CAH: cos(28°) = 9/Hyp
Hyp = 9 / cos(28°) = 9 / 0.8829
Hyp = 10.2 m

Q3

Find angle θ when the opposite = 5 m and hypotenuse = 13 m.

[2]

SOH: sin θ = 5/13 = 0.3846
θ = sin⁻¹(0.3846)
θ = 22.6°

Q4

Find angle θ when opposite = 8 m and adjacent = 6 m.

[2]

TOA: tan θ = 8/6 = 1.3333
θ = tan⁻¹(1.3333)
θ = 53.1°

Q5

A slope rises 3 m over a horizontal distance of 12 m. Find the angle of the slope.

[3]

OPP=3, ADJ=12 → TOA
tan θ = 3/12 = 0.25
θ = tan⁻¹(0.25)
θ = 14.0°

🟡 Tier 2 — Word Problems

Q6

A conveyor belt is inclined at 22° to the horizontal. The belt is 15 m long. How high does it rise?

[3]

HYP=15m, angle=22°, OPP=rise
SOH: sin(22°) = rise/15
rise = 15 × sin(22°) = 15 × 0.3746
rise = 5.62 m

Q7

A cable is attached to the top of a 12 m vertical mast and anchored to the ground 9 m from the base. Find the angle the cable makes with the ground.

[3]

OPP=12m (mast height), ADJ=9m (ground distance) → TOA
tan θ = 12/9 = 1.3333
θ = tan⁻¹(1.3333)
θ = 53.1°

Q8

From a point 80 m from the base of a building, the angle of elevation to the top is 38°. Find the height of the building.

[3]

ADJ=80m, angle=38°, OPP=height → TOA
tan(38°) = h/80
h = 80 × tan(38°) = 80 × 0.7813
h = 62.5 m

Q9

A roof has a pitch angle of 30° and a half-span of 6 m. Find the rafter length and the ridge height.

[4]

ADJ=6m (half-span), angle=30°
Rafter (HYP): cos(30°)=6/H → H=6/cos(30°)=6/0.8660 = 6.93 m
Ridge (OPP): tan(30°)=h/6 → h=6×tan(30°)=6×0.5774 = 3.46 m

Q10

A force of 350 N acts at 48° above the horizontal. Find the horizontal and vertical components.

[4]

Fₕ = 350 × cos(48°) = 350 × 0.6691 = 234.2 N
Fᵥ = 350 × sin(48°) = 350 × 0.7431 = 260.1 N

🔴 Tier 3 — T Level Challenge

Q11

A wheelchair ramp must meet the gradient 1:15 (rise 1 for every 15 horizontal). Find the angle and the ramp length if the rise is 0.8 m.

[4]

OPP=0.8m, ADJ=15×0.8=12m
tan θ = 0.8/12 = 0.0667 → θ = tan⁻¹(0.0667) = 3.8°
Ramp: H = 0.8/sin(3.8°) = 0.8/0.0663 = 12.07 m

Q12

A ship sails due east for 12 km, then on a bearing of N 40° E for 8 km. Using the right triangle formed, find the final distance north and east from the start.

[4]

Second leg: HYP=8km, angle=40° from north (so 50° from east)
North component = 8×cos(40°) = 8×0.766 = 6.13 km
East component = 8×sin(40°) = 8×0.643 = 5.14 km
Total east = 12 + 5.14 = 17.14 km east
Total north = 6.13 km north

Q13

A lathe taper has a large diameter of 60 mm, small diameter of 36 mm, and a length of 90 mm. Find the taper angle (half-angle of the taper).

[3]

Half-difference of radii = (60−36)/2 = 12 mm over length 90 mm
tan θ = 12/90 = 0.1333
θ = tan⁻¹(0.1333)
θ = 7.6°

Q14

From the top of a 50 m tower, the angles of depression to two points A and B on the same horizontal line are 35° and 20°. Both points are on the same side. Find the distance AB.

[5]

Distance to A: dₐ = 50/tan(35°) = 50/0.7002 = 71.4 m
Distance to B: d₃ = 50/tan(20°) = 50/0.3640 = 137.4 m
AB = 137.4 − 71.4 = 66.0 m ✅

Q15

A roof truss forms a right triangle. The horizontal tie is 9.6 m, the rafter makes 34° with the horizontal. Find: (a) the rafter length, (b) the vertical height, (c) the area of the triangular truss cross-section.

[6]

(a) Rafter: cos(34°)=9.6/H → H=9.6/cos(34°)=9.6/0.829 = 11.58 m
(b) Height: tan(34°)=h/9.6 → h=9.6×0.6745 = 6.48 m
(c) Area = ½×base×height = ½×9.6×6.48 = 31.1 m²

🎯 Score guide: Q1–5 = foundations (12 marks) · Q6–10 = word problems (17 marks) · Q11–15 = T Level challenge (22 marks)

👉 Next topic: Pythagoras’ Theorem in context — and the Sine & Cosine Rules for non-right-angled triangles.