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Leaving Certificate • Ordinary Level

Trigonometry

Exam-focused notes, key formulas, common mistakes, and interactive practice cards (with solutions).

Overview Formulas Question Types Common Mistakes Practice Checklist

Overview: what OL trigonometry usually tests Notes
  • Solving right-angled triangles using \(\sin\), \(\cos\), \(\tan\).
  • Using Pythagoras’ theorem in right-angled triangle contexts.
  • Angles of elevation and depression in real-life diagrams.
  • Solving non-right-angled triangles using the sine rule and cosine rule.
  • Area of a triangle using \(\tfrac{1}{2}ab\sin C\) in multi-step problems.
Calculator: Set your calculator to DEG (degrees). Rounding instructions matter.
Key formulas (must-know) Formulas

Right-angled triangles (SOHCAHTOA)

\[ \sin\theta=\frac{\text{opposite}}{\text{hypotenuse}},\quad \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}},\quad \tan\theta=\frac{\text{opposite}}{\text{adjacent}} \]

Pythagoras’ theorem

\[ a^2+b^2=c^2 \]

Cosine rule (non-right-angled triangles)

\[ a^2=b^2+c^2-2bc\cos A \]

Sine rule (non-right-angled triangles)

\[ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} \]

Area of a triangle using sine

\[ \text{Area}=\tfrac12 ab\sin C \]
Question types (what to recognise quickly) Exam Skills

1) Right-angled triangle (find a side)

  • Given: one angle + one side
  • Find: another side
Method: label opposite/adjacent/hypotenuse → choose ratio → solve.

2) Right-angled triangle (find an angle)

  • Given: two sides
  • Find: an angle
Method: form a ratio → use \(\sin^{-1}\), \(\cos^{-1}\), or \(\tan^{-1}\).

3) Elevation / depression

  • Often a “line of sight” diagram
  • Height and distance from a horizontal
Tip: draw a horizontal line; use alternate angles where needed.

4) Pythagoras then trig

  • Part (a): find a missing side using \(a^2+b^2=c^2\)
  • Part (b): use trig ratios with the result
Tip: keep intermediate answers to 3 significant figures (unless told otherwise).

5) Cosine rule

  • Non-right-angled triangle
  • Two sides and the included angle known
Tip: watch brackets: \(b^2+c^2-2bc\cos A\).

6) Sine rule

  • Non-right-angled triangle
  • At least one opposite side–angle pair known
Tip: always match a side with its opposite angle.

7) Area using \(\tfrac12 ab\sin C\)

  • Two sides and the included angle given
  • Often embedded in longer questions
Tip: include correct units (e.g. \(\text{m}^2\)).

8) Multi-step “real-life” problem

  • Mix of Pythagoras + trig + sine/cosine rule
  • Usually high-value marks
Tip: answer parts in order; method marks are key at OL.
Common mistakes (avoid losing easy marks) Warnings
  • Calculator not in DEG mode.
  • Mixing up opposite and adjacent.
  • Using sine/cosine rule in a right-angled triangle.
  • Forgetting units for length/area.
  • Ignoring rounding instructions (nearest degree / 1 d.p. / nearest metre).
Practice: exam-style questions (with reveal solutions) Interactive

Question 1

In a right-angled triangle, the hypotenuse is \(12\text{ cm}\) and an acute angle is \(35^\circ\). Find the length of the side opposite the \(35^\circ\) angle. Give your answer correct to 1 decimal place.

Solution

\[ \sin 35^\circ=\frac{\text{opposite}}{12} \Rightarrow \text{opposite}=12\sin 35^\circ \] \[ \text{opposite}\approx 12(0.5736)=6.8832\approx 6.9\text{ cm} \]

Marking scheme (typical)

  • Correct ratio set up: 2
  • Correct substitution: 1
  • Correct calculation: 1
  • Correct rounding + units: 1

Question 2

From a point on level ground, the angle of elevation to the top of a building is \(28^\circ\). The point is \(40\text{ m}\) from the base of the building. Find the height of the building, correct to the nearest metre.

Solution

\[ \tan 28^\circ=\frac{h}{40}\Rightarrow h=40\tan 28^\circ \] \[ h\approx 40(0.5317)=21.268\approx 21\text{ m} \]

Marking scheme (typical)

  • Correct ratio: 2
  • Correct rearrangement: 1
  • Correct calculation: 1
  • Nearest metre + units: 1

Question 3

In triangle \(ABC\), \(AB=9\text{ cm}\), \(AC=7\text{ cm}\), and \(\angle A=52^\circ\). Find \(BC\), correct to 1 decimal place.

Solution

Use cosine rule: \[ BC^2=9^2+7^2-2(9)(7)\cos 52^\circ \] \[ BC^2=81+49-126\cos 52^\circ \] \[ BC^2\approx 130-126(0.6157)=130-77.5782=52.4218 \] \[ BC\approx \sqrt{52.4218}=7.24\approx 7.2\text{ cm} \]

Marking scheme (typical)

  • Correct formula: 2
  • Correct substitution: 1
  • Correct arithmetic: 1
  • Square root, rounding + units: 1

Question 4

Two sides of a triangle are \(a=11\text{ m}\) and \(b=8\text{ m}\) and the included angle is \(39^\circ\). Find the area of the triangle, correct to the nearest square metre.

Solution

\[ \text{Area}=\tfrac12 ab\sin C=\tfrac12(11)(8)\sin 39^\circ \] \[ \text{Area}=44\sin 39^\circ\approx 44(0.6293)=27.689\approx 28\text{ m}^2 \]

Marking scheme (typical)

  • Correct formula: 2
  • Correct substitution: 1
  • Correct calculation: 1
  • Nearest \(\text{m}^2\) + units: 1
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