Trig Ratios for Any Angle — Grade 10 Maths (CAPS)

What sin, cos and tan mean once the angle grows past 90° — the x, y and r definitions, the CAST diagram for signs, reference angles, reading every ratio off a point, working backwards from one ratio, the reciprocals cosec, sec and cot, and a Grade 11 preview of solving a trig equation across 0° to 360°.

Also known as: CAST diagram · reference angle · trig ratios any angle · quadrants trigonometry · cosec sec cot

SOH-CAH-TOA only works inside a right-angled triangle, so it only handles angles between 0° and 90°. But a Paper 2 question will happily ask for sin 210° or tell you that cosθ is negative. This lesson extends the three ratios to every angle from 0° to 360° — and the method here is the foundation of all the trigonometry in Grade 11 and 12.

1Step 1: put the angle on the Cartesian plane

Instead of a triangle floating on its own, draw the angle in standard position: its vertex at the origin, one arm along the positive x-axis, and the other arm rotating anticlockwise. Pick any point P(x ; y) on that rotating arm.

Drop a perpendicular from P to the x-axis and you get a right-angled triangle again — but now with signed coordinates.

The angle θ is measured anticlockwise from the positive x-axis. β is the reference angle — the acute angle between the arm and the nearest part of the x-axis.

r = √(x² + y²) and r is always positive

The three ratios are now defined by coordinates, not by "opposite" and "adjacent":

sin θ = y/r cos θ = x/r tan θ = y/x (x ≠ 0)

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These are not new rules. In the first quadrant y is the opposite side, x is the adjacent side and r is the hypotenuse — so they reduce to SOH-CAH-TOA exactly. All that changes is that x and y may now be negative.

2Step 2: the signs — the CAST diagram

Because r is always positive, the sign of each ratio depends only on the signs of x and y. In the second quadrant, for instance, x is negative and y is positive, so cosθ = x/r is negative while sinθ = y/r is positive.

Working that out every time is slow. CAST records the answer: each letter names the ratio that is positive in that quadrant.

Read CAST anticlockwise from the fourth quadrant: C, A, S, T. Each letter marks the ratio that is positive there — everything else is negative.

Quadrant 1 (0°–90°) — All three are positive.
Quadrant 2 (90°–180°) — only Sin is positive.
Quadrant 3 (180°–270°) — only Tan is positive.
Quadrant 4 (270°–360°) — only Cos is positive.

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Do not read CAST clockwise from the top. It starts in the fourth quadrant with C and runs anticlockwise: C → A → S → T. Reading it the wrong way round swaps sine and cosine and quietly costs you every sign in the question.

3Step 3: the reference angle

The reference angle β is the acute angle between the rotating arm and the x-axis — never the y-axis. It is what your calculator gives you when you use sin⁻¹, cos⁻¹ or tan⁻¹, because a calculator only ever returns an acute answer for a positive ratio.

Q1: θ = β Q2: θ = 180° − β Q3: θ = 180° + β Q4: θ = 360° − β

The size of the ratio comes from β; the sign comes from CAST. That is the whole method.

Worked ExampleEvaluate sin 210° without a calculator

1
210° lies between 180° and 270°, so it is in Quadrant 3.
Identify the quadrant first — it fixes the sign.
2
β = 210° − 180° = 30°
In Q3 the reference angle is θ − 180°.
3
In Q3 only tan is positive, so sin 210° is negative.
CAST.
4
sin 210° = −sin 30° = −½
Size from the reference angle, sign from CAST.

Worked ExampleEvaluate cos 315° and tan 120°

1
315° is in Q4, so β = 360° − 315° = 45°. In Q4 cos is positive.
2
cos 315° = +cos 45° = (√2)/2
3
120° is in Q2, so β = 180° − 120° = 60°. In Q2 only sin is positive, so tan is negative.
4
tan 120° = −tan 60° = −√3

4Reading every ratio off a point

If a question gives you a point, you never need a calculator — you have x and y, and Pythagoras gives r.

Worked ExampleP(−3 ; 4) lies on the arm of θ. Determine sin θ, cos θ and tan θ.

1
x = −3, y = 4
Read them straight off the point — keep the signs.
2
r = √((−3)² + 4²) = √(9 + 16) = √25 = 5
r is a length, so it is always positive.
3
sin θ = y/r = 4/5
4
cos θ = x/r = (−3)/5
Negative, because the point is to the left of the y-axis.
5
tan θ = y/x = 4/(−3) = (−4)/3
6
Check: P is in Q2, where only sin is positive. Sine came out positive; cosine and tangent negative. ✓
Always sanity-check against CAST — it catches a dropped minus.

5Working backwards from one ratio

The harder version gives you one ratio plus a condition, and asks for the others. Do not reach for the calculator — sketch the point instead.

Worked ExampleIf sin θ = 5/13 and cos θ < 0, determine cos θ and tan θ.

1
sin θ is positive and cos θ is negative → Quadrant 2.
Two sign conditions pin the quadrant down.
2
sin θ = y/r, so y = 5 and r = 13.
Match the ratio to the definition.
3
x² = r² − y² = 169 − 25 = 144, so x = ±12.
Pythagoras.
4
In Q2, x is negative, so x = −12.
The quadrant decides which root to take — this is the step most learners skip.
5
cos θ = x/r = (−12)/13 and tan θ = y/x = (−5)/12

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Taking x = +12 because Pythagoras gave a positive answer. Pythagoras gives you the length; the quadrant tells you the sign. Half the marks in this question type are for getting that minus right.

6The reciprocal ratios: cosec, sec and cot

Each ratio has a reciprocal, and they are examined in Grade 10.

cosec θ = 1/(sin θ) = r/y sec θ = 1/(cos θ) = r/x cot θ = 1/(tan θ) = x/y
Each is undefined where its denominator is 0: cosec and cot need y ≠ 0, sec needs x ≠ 0.

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The pairing is deliberately awkward: cosec goes with sin and sec goes with cos — not the other way round. Match them by the third letter: cosec → sin, sec → cos.

A reciprocal has the same sign as the ratio it comes from, so CAST covers all six. And since you cannot divide by zero, cosecθ is undefined wherever sinθ = 0 (at 0°, 180°, 360°) and secθ is undefined at 90° and 270°.

7Looking ahead: Grade 11 — solving over 0° to 360°

Everything above is Grade 10 work. Solving a trig equation is not, once the interval runs past 90°: in Grade 10 you solve simple trig equations only for angles between 0° and 90° (that is the lesson Special Angles, Reciprocal Ratios & Trig Equations). The rest of this section is the Grade 11 step, put here so you can see where CAST and the reference angle are leading. It is not examined in Grade 10, and nothing below it is set in this lesson's practice, quiz or paper.

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Straight from the ATP. Grade 10: “Solve simple trigonometric equations for angles between 0° and 90°” and “Use a diagram to determine the numerical values of ratios for angles from 0° to 360°” — so evaluating any ratio in 0°–360° is yours now; solving across that range is not. Grade 11: “Determine the general solutions of trigonometric equations. Also, determine solutions in specific intervals.”

Now the pieces combine. A question that asks for all solutions in an interval wants one answer per quadrant where the sign works.

Worked ExampleSolve 2 sin θ + 1 = 0 for θ ∈ [0° ; 360°]

1
2 sin θ = −1, so sin θ = −½
Isolate the ratio first — always.
2
Reference angle: sin β = ½, so β = 30°.
Use the POSITIVE value to find β. The calculator only returns acute angles.
3
sin θ is negative in Q3 and Q4.
CAST: sin is positive only in Q1 and Q2.
4
Q3: θ = 180° + 30° = 210°
5
Q4: θ = 360° − 30° = 330°
6
θ = 210° or θ = 330°
Two solutions in the interval — a single answer would lose half the marks.

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Typing sin⁻¹(−0,5) into a calculator returns −30°, which is not in the interval, and stopping there gives no valid answer at all. Find β from the positive ratio, then place it in the correct quadrants yourself.

8The Grade 11 routine, every time

Isolate the ratio (get sin θ = … on its own).
Find the reference angle β from the positive value.
Use CAST to decide which quadrants the sign allows.
Convert β to θ in each of those quadrants: β, 180° − β, 180° + β, 360° − β.
Check every answer falls inside the interval you were given.

Practice exercises

Work each one out, then click to reveal the answer.

📄 Worksheet ✅ Memo 📝 Exam paper 🗝️ Marking guide

1
Write down the quadrant in which 235° lies, and state which of sin, cos and tan is positive there.
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Quadrant 3; only tan is positive.
2
Determine the reference angle for θ = 295°.
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Q4, so β = 360° − 295° = 65°
3
Evaluate cos 150° without using a calculator.
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Q2, β = 30°, cos is negative in Q2, so cos 150° = −cos 30° = −(√3)/2
4
Evaluate tan 225° without using a calculator.
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Q3, β = 45°, tan is positive in Q3, so tan 225° = +tan 45° = 1
5
P(−8 ; −6) lies on the arm of θ. Determine sinθ, cosθ and tanθ.
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r = √(64 + 36) = 10 ✓; sinθ = −6/10 = −3/5 ✓; cosθ = −8/10 = −4/5 ✓; tanθ = −6/−8 = 3/4 ✓ (Q3, so only tan is positive)
6
If cosθ = 8/17 and sinθ < 0, determine sinθ and tanθ.
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cos positive, sin negative → Q4. x = 8, r = 17, y² = 289 − 64 = 225, y = ±15; in Q4 y is negative so y = −15. sinθ = −15/17, tanθ = −15/8
7
Write down the value of cosec 30°.
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cosec 30° = 1 ÷ sin 30° = 1 ÷ ½ = 2
8
Explain why sec 90° is undefined.
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secθ = 1/cosθ and cos 90° = 0; division by zero is undefined.
9
Evaluate cosec 210° without using a calculator.
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210° is in Q3, so β = 210° − 180° = 30°; in Q3 only tan is positive, so sin 210° = −sin 30° = −½; cosec 210° = 1 ÷ (−½) = −2
10
Without using a calculator, show that sin 150° + cos 300° = 1.
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sin 150°: Q2, β = 180° − 150° = 30°, and sin is positive in Q2, so sin 150° = +sin 30° = ½; cos 300°: Q4, β = 360° − 300° = 60°, and cos is positive in Q4, so cos 300° = +cos 60° = ½; ½ + ½ = 1