Grade 10 Mathematics
Grade 10 · Term 1Mathematics

Trigonometry

Trigonometry ('triangle measurement') links angles to side-length ratios. In Term 1 we define sin, cos and tan for right-angled triangles, extend these definitions to all angles from 0° to 360° using the Cartesian plane (CAST rule), master the exact values at special angles without a calculator, and apply everything to two-dimensional problems.

3.1 Trigonometric Ratios3.2 Special Angles, Extended Definitions & CAST3.3 Solving Trig Equations & 2D Problems
Week 9

3.1 Trigonometric Ratios

  • Define the trigonometric ratios sin θ, cos θ and tan θ using a right-angled triangle
  • Define the reciprocal ratios: cosec θ, sec θ, cot θ
  • Use the ratios to find unknown sides and angles in right-angled triangles
🌍

Real-World Connection

A surveyor measuring a building from a fixed distance uses the tangent ratio every time: $\tan(\text{angle of elevation}) = \frac{\text{height}}{\text{distance}}$. The angle and one known side are enough to find any other measurement — that is the power of trigonometric ratios.

Definition

Hypotenuse, Opposite, Adjacent

In a right-angled triangle, the hypotenuse (hyp) is always opposite the right angle and is the longest side. The opposite (opp) and adjacent (adj) sides depend on which angle θ is your reference point.

hyp=longest side,opp=side facing θ,adj=side next to θ\text{hyp} = \text{longest side},\quad \text{opp} = \text{side facing } \theta,\quad \text{adj} = \text{side next to } \theta

Sine ratio

sin⁡θ=opphyp\sin\theta = \dfrac{\text{opp}}{\text{hyp}}

θ is the reference angle

Cosine ratio

cos⁡θ=adjhyp\cos\theta = \dfrac{\text{adj}}{\text{hyp}}

θ is the reference angle

Tangent ratio

tan⁡θ=oppadj\tan\theta = \dfrac{\text{opp}}{\text{adj}}

θ is the reference angle; note $\tan\theta=\frac{\sin\theta}{\cos\theta}$

💡 Tip

Memory aid: SOH-CAH-TOA. Sin=Opp/Hyp, Cos=Adj/Hyp, Tan=Opp/Adj.

Reciprocal ratios (examined in Grade 10 only)

  • cosec⁡θ=1sin⁡θ=hypopp\cosec\theta = \dfrac{1}{\sin\theta} = \dfrac{\text{hyp}}{\text{opp}}
  • sec⁡θ=1cos⁡θ=hypadj\sec\theta = \dfrac{1}{\cos\theta} = \dfrac{\text{hyp}}{\text{adj}}
  • cot⁡θ=1tan⁡θ=adjopp\cot\theta = \dfrac{1}{\tan\theta} = \dfrac{\text{adj}}{\text{opp}}

🚨 Common Mistake

sin⁡−1\sin^{-1} on a calculator is NOT the same as cosec⁡\cosec. The inverse function sin⁡−1\sin^{-1} gives you the angle; cosec⁡θ=1/sin⁡θ\cosec\theta=1/\sin\theta gives you a ratio value.

Worked Examples

Worked Example

Find all six trigonometric ratios

Problem

In △DEF\triangle DEF, E^=90°\hat{E}=90°, DE=5DE=5, EF=12EF=12. Find sin⁡θ\sin\theta, cos⁡θ\cos\theta, tan⁡θ\tan\theta, cosec⁡θ\cosec\theta, sec⁡θ\sec\theta, cot⁡θ\cot\theta where θ=F^\theta=\hat{F}.

Worked Example

Find an unknown side using a ratio

Problem

In △ABC\triangle ABC, C^=90°\hat{C}=90°, A^=35°\hat{A}=35°, AB=10AB=10 cm. Find BCBC correct to 2 decimal places.
Activity — 8 Questions

CAPS Cognitive Level Distribution

L1 · Knowledge2 Q
L2 · Routine Procedures2 Q
L3 · Complex Procedures2 Q
L4 · Problem Solving2 Q
1
L1 · Knowledge2 marks
State the SOH-CAH-TOA definition for tan⁡θ\tan\theta.
2
L1 · Knowledge2 marks
In a right triangle, sin⁡θ=725\sin\theta = \dfrac{7}{25}. Write down cosec⁡θ\cosec\theta.
3
L2 · Routine Procedures4 marks
In △PQR\triangle PQR, R^=90°\hat{R}=90°, PQ=17PQ=17, QR=8QR=8. Find sin⁡Q^\sin\hat{Q}, cos⁡Q^\cos\hat{Q}, tan⁡Q^\tan\hat{Q}.
4
L2 · Routine Procedures4 marks
In a right triangle, cos⁡α=513\cos\alpha=\dfrac{5}{13}. Find sin⁡α\sin\alpha and tan⁡α\tan\alpha without using a calculator.
5
L3 · Complex Procedures4 marks
In △ABC\triangle ABC, C^=90°\hat{C}=90°, tan⁡A=34\tan A=\dfrac{3}{4}. Find sin⁡B\sin B and cos⁡A\cos A.
6
L3 · Complex Procedures3 marks
Find xx in cm correct to 2 d.p. if sin⁡52°=x20\sin52°=\dfrac{x}{20}.
7
L4 · Problem Solving5 marks
A ladder 8 m long leans against a wall. It makes a 65° angle with the ground. How high up the wall does the ladder reach? Give your answer correct to 1 decimal place.
8
L4 · Problem Solving4 marks
Show that sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1 using the sides of a right-angled triangle with hypotenuse rr and legs xx, yy.
Week 10

3.2 Special Angles, Extended Definitions & CAST

  • Extend definitions of sin θ, cos θ, tan θ for 0° ≤ θ ≤ 360° using the Cartesian plane
  • Derive values for special angles 0°, 30°, 45°, 60°, 90° without a calculator
  • Use the CAST diagram to determine signs in each quadrant
🌍

Real-World Connection

Imagine standing at the centre of a circular clock face and rotating a hand anti-clockwise. At any position the hand makes an angle θ with the positive x-axis (3 o'clock). The y-coordinate of the tip is $r\sin\theta$ and the x-coordinate is $r\cos\theta$ — these ratios work for ALL angles, not just acute ones.

Definition

Extended definitions using coordinates

For any point R(x;y)R(x; y) on a circle of radius rr centred at the origin, with angle θ\theta in standard position (measured anti-clockwise from the positive x-axis):

sin⁡θ=yr,cos⁡θ=xr,tan⁡θ=yx  (x≠0)\sin\theta = \frac{y}{r},\quad \cos\theta = \frac{x}{r},\quad \tan\theta = \frac{y}{x}\;(x\neq0)

Property / Rule

Signs in each quadrant — CAST rule

Only one ratio is positive per quadrant (except Q1 where all are positive). The word CAST (starting from Q4, going anti-clockwise) summarises which is positive in each quadrant.

Q1: All  ∣  Q2: Sin  ∣  Q3: Tan  ∣  Q4: Cos\text{Q1: All}\;|\;\text{Q2: Sin}\;|\;\text{Q3: Tan}\;|\;\text{Q4: Cos}

Definition

Special angle values

Derived from two standard triangles: 45°-45°-90° (isoceles right triangle, legs = 1) and 30°-60°-90° (half an equilateral triangle, side = 2).

See table below\text{See table below}

Special angle exact values

  • θ=0°\theta=0°: sin⁡0°=0\sin0°=0, cos⁡0°=1\cos0°=1, tan⁡0°=0\tan0°=0
  • θ=30°\theta=30°: sin⁡30°=12\sin30°=\tfrac{1}{2}, cos⁡30°=32\cos30°=\tfrac{\sqrt{3}}{2}, tan⁡30°=13=33\tan30°=\tfrac{1}{\sqrt{3}}=\tfrac{\sqrt{3}}{3}
  • θ=45°\theta=45°: sin⁡45°=12=22\sin45°=\tfrac{1}{\sqrt{2}}=\tfrac{\sqrt{2}}{2}, cos⁡45°=22\cos45°=\tfrac{\sqrt{2}}{2}, tan⁡45°=1\tan45°=1
  • θ=60°\theta=60°: sin⁡60°=32\sin60°=\tfrac{\sqrt{3}}{2}, cos⁡60°=12\cos60°=\tfrac{1}{2}, tan⁡60°=3\tan60°=\sqrt{3}
  • θ=90°\theta=90°: sin⁡90°=1\sin90°=1, cos⁡90°=0\cos90°=0, tan⁡90°=\tan90°= undefined

💡 Tip

For the sin column, read downwards: 0,12,12,32,10,\tfrac{1}{2},\tfrac{1}{\sqrt{2}},\tfrac{\sqrt{3}}{2},1 — the values increase from 0 to 1. The cos column is the sin column reversed.

Worked Examples

Worked Example

Evaluate without a calculator

Problem

Calculate: (a) sin⁡30°+cos⁡60°\sin30°+\cos60°(b) \quad (b) tan⁡45°⋅cos⁡30°−sin⁡60°\tan45°\cdot\cos30°-\sin60°

Worked Example

Use CAST to find a ratio

Problem

If cos⁡θ=−45\cos\theta = -\dfrac{4}{5} and θ∈[90°;270°]\theta\in[90°;270°], find sin⁡θ\sin\theta and tan⁡θ\tan\theta without a calculator.
Activity — 8 Questions

CAPS Cognitive Level Distribution

L1 · Knowledge2 Q
L2 · Routine Procedures2 Q
L3 · Complex Procedures2 Q
L4 · Problem Solving2 Q
1
L1 · Knowledge1 mark
State the exact value of sin⁡45°\sin45°.
2
L1 · Knowledge2 marks
In which quadrants is tan⁡θ\tan\theta positive?
3
L2 · Routine Procedures3 marks
Evaluate without a calculator: cos⁡30°+tan⁡60°\cos30°+\tan60°.
4
L2 · Routine Procedures2 marks
In which quadrant does θ\theta lie if sin⁡θ>0\sin\theta>0 and cos⁡θ<0\cos\theta<0?
5
L3 · Complex Procedures5 marks
If sin⁡θ=−513\sin\theta=-\dfrac{5}{13} and θ∈(180°;360°)\theta\in(180°;360°), find cos⁡θ\cos\theta and tan⁡θ\tan\theta.
6
L3 · Complex Procedures3 marks
Evaluate without a calculator: sin⁡60°tan⁡30°\dfrac{\sin60°}{\tan30°}.
7
L4 · Problem Solving5 marks
Without a calculator, evaluate tan⁡230°+sin⁡245°+cos⁡260°\tan^2 30°+\sin^2 45°+\cos^2 60° and express the answer as a fraction.
8
L4 · Problem Solving3 marks
For which value(s) of θ∈[0°;360°]\theta\in[0°;360°] is tan⁡θ\tan\theta undefined? Give a geometric reason.
Week 11

3.3 Solving Trig Equations & 2D Problems

  • Solve simple trigonometric equations for angles between 0° and 90°
  • Solve two-dimensional problems involving right-angled triangles
🌍

Real-World Connection

A ship's navigator uses right-angle triangles to find position: given a bearing and distance, the north and east displacements are the two legs of a right triangle. Solving for the angle is the reverse — it is the same equation, just with the unknown on the other side.

Property / Rule

Solving a trig equation for acute angles

Isolate the trig ratio, then use the inverse function (or the special-angle table) to find the angle. For acute angles, there is exactly one solution.

sin⁡θ=12⇒θ=sin⁡−1(12)=30°\sin\theta = \tfrac{1}{2} \Rightarrow \theta = \sin^{-1}(\tfrac{1}{2}) = 30°

Property / Rule

Angles of elevation and depression

Angle of elevation: measured upward from the horizontal. Angle of depression: measured downward from the horizontal. Both are always positive acute angles.

tan⁡(elevation angle)=vertical risehorizontal distance\tan(\text{elevation angle}) = \dfrac{\text{vertical rise}}{\text{horizontal distance}}

💡 Tip

Always draw a clear diagram for 2D problems. Label the known sides and angles, identify which trig ratio connects the known and unknown, then write the equation before solving.

Worked Examples

Worked Example

Solve a simple trigonometric equation

Problem

Solve for xx where 0°≤x≤90°0°\leq x\leq90°: (a) 2cos⁡x=32\cos x = \sqrt{3}(b) \quad (b) tan⁡x−1=0\tan x - 1 = 0

Worked Example

Two-dimensional problem

Problem

From a point AA on level ground, the angle of elevation to the top TT of a building is 52°. The base of the building BB is 30 m from AA. Calculate the height BTBT correct to 1 decimal place.
Activity — 8 Questions

CAPS Cognitive Level Distribution

L1 · Knowledge2 Q
L2 · Routine Procedures2 Q
L3 · Complex Procedures2 Q
L4 · Problem Solving2 Q
1
L1 · Knowledge2 marks
Solve for θ\theta: sin⁡θ=0,5\sin\theta = 0{,}5, where 0°≤θ≤90°0°\leq\theta\leq90°.
2
L1 · Knowledge2 marks
State the angle of elevation if tan⁡θ=1\tan\theta=1 for an acute angle.
3
L2 · Routine Procedures3 marks
Solve for xx where 0°≤x≤90°0°\leq x\leq90°: 2sin⁡x−2=02\sin x - \sqrt{2}=0.
4
L2 · Routine Procedures4 marks
A ramp rises 1{,}2 m over a horizontal distance of 5 m. Find the angle the ramp makes with the ground correct to 1 decimal place.
5
L3 · Complex Procedures3 marks
Solve for α\alpha: 3tan⁡α−3=03\tan\alpha-\sqrt{3}=0 where α\alpha is acute. Give exact answer.
6
L3 · Complex Procedures5 marks
From the top of a cliff 80 m high, the angle of depression to a boat is 35°. How far is the boat from the base of the cliff? Give answer correct to 1 decimal place.
7
L4 · Problem Solving6 marks
Two buildings stand on level ground. From the base of the shorter building (height 20 m), the angle of elevation to the top of the taller building is 40°. The buildings are 35 m apart. Find the height of the taller building correct to 1 decimal place.
8
L4 · Problem Solving6 marks
A triangular plot ABCABC has B^=90°\hat{B}=90°, AB=24AB=24 m and BC=7BC=7 m. A fence post PP is placed on ACAC such that BP⊥ACBP\perp AC. Find the length BPBP correct to 2 decimal places.

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Exponents, Equations & Inequalities

Trigonometry Grade 10 Maths CAPS Notes & Examples | MathSciBuddy