Geometry and Measurement Sec 2 E-Mathematics

Trigonometry

Trigonometry Study Notes

Key Concepts

  • Trigonometry is a branch of Mathematics that deals with the relationship between the angles and sides of triangles.
  • At this level, trigonometry is mainly used with right-angled triangles.

1. Right-angled triangle

  • A right-angled triangle is a triangle with one angle equal to 90°.
  • The side opposite the 90° angle is always the hypotenuse.
  • The hypotenuse is the longest side in the triangle.

2. Naming the sides of a right-angled triangle

To use trigonometry, you must know how to identify the sides relative to a chosen angle.

  • Hypotenuse
    • The side opposite the right angle.
  • Opposite side
    • The side directly opposite the angle you are focusing on.
  • Adjacent side
    • The side next to the angle you are focusing on, but it is not the hypotenuse.

Important: The opposite and adjacent sides can change depending on which angle you choose.

3. Trigonometric ratios

The three basic trigonometric ratios are:

  • Sine
  • Cosine
  • Tangent

These ratios compare the lengths of sides in a right-angled triangle.

For an angle θ \theta :

  • sinθ=oppositehypotenuse \sin \theta = \dfrac{\text{opposite}}{\text{hypotenuse}}
  • cosθ=adjacenthypotenuse \cos \theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}
  • tanθ=oppositeadjacent \tan \theta = \dfrac{\text{opposite}}{\text{adjacent}}

4. SOH-CAH-TOA

This is a memory aid to help remember the trigonometric ratios.

  • SOHSine = Opposite / Hypotenuse
  • CAHCosine = Adjacent / Hypotenuse
  • TOATangent = Opposite / Adjacent

5. Finding unknown sides in right-angled triangles

To find an unknown side:

  1. Identify the given angle.
  2. Label the sides as opposite, adjacent, and hypotenuse.
  3. Choose the correct trigonometric ratio.
  4. Substitute the known values into the formula.
  5. Solve for the unknown side.
  6. Write the answer with the correct unit.

Example of choosing the ratio:

  • If you know opposite and want hypotenuse, use sine.
  • If you know adjacent and want hypotenuse, use cosine.
  • If you know opposite and want adjacent, use tangent.

6. Finding unknown angles using inverse trigonometric ratios

Sometimes the side lengths are known, and you need to find an angle.

Use the inverse trig functions on the calculator:

  • θ=sin1(oppositehypotenuse) \theta = \sin^{-1}\left(\dfrac{\text{opposite}}{\text{hypotenuse}}\right)
  • θ=cos1(adjacenthypotenuse) \theta = \cos^{-1}\left(\dfrac{\text{adjacent}}{\text{hypotenuse}}\right)
  • θ=tan1(oppositeadjacent) \theta = \tan^{-1}\left(\dfrac{\text{opposite}}{\text{adjacent}}\right)

7. Calculator use

  • Make sure the calculator is in degree mode when working with angles in degrees.
  • In triangle questions at this level, angles are usually given in degrees (°).

8. Angles of elevation and depression

These are used in real-life applications involving heights and distances.

Angle of elevation

  • The angle between the horizontal line and the line of sight when looking upwards.

Angle of depression

  • The angle between the horizontal line and the line of sight when looking downwards.

Key idea:

  • Horizontal lines are parallel, so the angle of depression from the top is equal to the angle of elevation from the bottom, if they are formed by the same line of sight.

9. Bearing problems

A bearing is a direction measured:

  • clockwise
  • from North
  • using 3 digits

Examples:

  • North-East direction may be written as 045°
  • East is 090°
  • South is 180°
  • West is 270°

When solving bearing problems:

  • Always draw a North line.
  • Bearings are measured clockwise from North.
  • Use trigonometry if a right-angled triangle can be formed.

10. When trigonometry is useful

Trigonometry can help you find:

  • heights of buildings
  • width of rivers
  • distance to objects
  • angles in navigation and map-reading
  • directions involving bearings

Important Definitions

  • Trigonometry: the study of the relationships between the sides and angles of triangles.
  • Right-angled triangle: a triangle with one angle of 90°.
  • Hypotenuse: the side opposite the right angle in a right-angled triangle; it is the longest side.
  • Opposite side: the side directly opposite the angle being considered.
  • Adjacent side: the side next to the angle being considered, excluding the hypotenuse.
  • Trigonometric ratio: a ratio comparing two sides of a right-angled triangle.
  • Sine: the ratio of the opposite side to the hypotenuse.
  • Cosine: the ratio of the adjacent side to the hypotenuse.
  • Tangent: the ratio of the opposite side to the adjacent side.
  • SOH-CAH-TOA: a mnemonic used to remember the definitions of sine, cosine and tangent.
  • Inverse trigonometric function: a function used to find an angle when the trigonometric ratio is known.
  • Angle of elevation: the angle between the horizontal and the line of sight when looking upwards.
  • Angle of depression: the angle between the horizontal and the line of sight when looking downwards.
  • Line of sight: the straight line from the observer’s eye to the object.
  • Bearing: the direction of one point from another, measured clockwise from North and written as a 3-digit angle.
  • Horizontal: a line parallel to the ground.
  • Vertical: a line perpendicular to the ground.

Worked Examples

Example 1: Finding an unknown side using sine

A ladder leans against a wall. The ladder is 5.0 m long and makes an angle of 40° with the ground. Find the height reached by the ladder on the wall.

Step 1: Identify the triangle

  • The ladder, wall and ground form a right-angled triangle.
  • The ladder is the hypotenuse.
  • The height up the wall is the opposite side to the 4040^\circ angle.

Step 2: Choose the correct ratio

We need opposite and hypotenuse, so use sine.

sin40=oppositehypotenuse \sin 40^\circ = \frac{\text{opposite}}{\text{hypotenuse}}
sin40=h5.0 \sin 40^\circ = \frac{h}{5.0}

Step 3: Solve

h=5.0sin40 h = 5.0 \sin 40^\circ
h=5.0×0.6428 h = 5.0 \times 0.6428
h=3.214 h = 3.214

Step 4: Write the answer

h3.21 m h \approx 3.21 \text{ m}

Answer: The ladder reaches a height of 3.21 m up the wall.


Example 2: Finding an unknown angle using inverse tangent

A tree casts a shadow of 8.0 m. The height of the tree is 6.0 m. Find the angle of elevation of the Sun.

Step 1: Identify the sides

  • Height of tree = opposite side = 6.0 m
  • Shadow length = adjacent side = 8.0 m

Step 2: Choose the correct ratio

We need opposite and adjacent, so use tangent.

tanθ=oppositeadjacent \tan \theta = \frac{\text{opposite}}{\text{adjacent}}
tanθ=6.08.0 \tan \theta = \frac{6.0}{8.0}
tanθ=0.75 \tan \theta = 0.75

Step 3: Use inverse tangent

θ=tan1(0.75) \theta = \tan^{-1}(0.75)
θ36.9 \theta \approx 36.9^\circ

Answer: The angle of elevation of the Sun is 36.936.9^\circ.


Example 3: Bearing and trigonometry

A boat sails 12 km due East from a lighthouse, then 5 km due North. Find:

  1. its distance from the lighthouse
  2. the bearing of the boat from the lighthouse

Step 1: Draw the path

  • Start at the lighthouse LL
  • Move 12 km East
  • Then move 5 km North to boat BB

This forms a right-angled triangle:

  • horizontal side = 12 km
  • vertical side = 5 km

Step 2: Find the distance

Use Pythagoras’ theorem:

LB=122+52 LB = \sqrt{12^2 + 5^2}
LB=144+25 LB = \sqrt{144 + 25}
LB=169 LB = \sqrt{169}
LB=13 LB = 13

So the boat is 13 km from the lighthouse.

Step 3: Find the angle

To find the bearing, first find the angle from the East direction or North direction.

Using angle from East:

tanθ=512 \tan \theta = \frac{5}{12}
θ=tan1(512) \theta = \tan^{-1}\left(\frac{5}{12}\right)
θ22.6 \theta \approx 22.6^\circ

This means the boat is 22.622.6^\circ north of East.

Step 4: Convert to bearing

Bearing is measured clockwise from North.

From North to East is 9090^\circ. Since the direction is 22.622.6^\circ above East:

bearing=9022.6=67.4 \text{bearing} = 90^\circ - 22.6^\circ = 67.4^\circ

Write as a 3-digit bearing:

067.4 \boxed{067.4^\circ}

If the question requires whole-number bearing:

067 \boxed{067^\circ}

Answers:

  1. Distance from lighthouse = 13 km
  2. Bearing of boat from lighthouse = 067° approximately

Bearings

Definition

A bearing is an angle used to describe direction. It is always:

  • Measured clockwise from North
  • Written as 3 digits (e.g. 045°, not 45°; 090°, not 90°)

Key Bearing Facts

Direction Bearing
North 000°
East 090°
South 180°
West 270°

Back Bearings (Reverse Bearings)

When you need to find the bearing of A from B, given the bearing of B from A, use:

Back bearing = bearing + 180°

If the result is greater than 360°, subtract 360°.

Worked Example:

“Town B is on a bearing of 120° from Town A. Find the bearing of A from B.”

Step 1: The bearing of B from A is 120°.

Step 2: Apply the back-bearing rule: [\text{Back bearing} = 120° + 180° = 300°]

Step 3: Check — 300° is less than 360°, so no further adjustment needed.

Answer: A is on a bearing of 300° from B.

Another example (result exceeds 360°):

  • Bearing of B from A = 250°
  • Back bearing = 250° + 180° = 430°
  • 430° > 360°, so subtract 360°: 430° − 360° = 070°
  • Bearing of A from B = 070°

Angles of Elevation and Depression (Recap)

These concepts appear alongside bearings in real-life trigonometry problems:

  • Angle of elevation: The angle measured upwards from the horizontal to a line of sight
    • Example: Looking up at the top of a building from ground level
  • Angle of depression: The angle measured downwards from the horizontal to a line of sight
    • Example: Looking down at a boat from the top of a cliff

Key property: The angle of elevation from point A to point B equals the angle of depression from point B to point A (alternate angles, parallel horizontal lines).


Common Mistakes to Avoid

  • Mixing up opposite, adjacent, and hypotenuse.
  • Forgetting that the hypotenuse is always opposite the 90° angle.
  • Using the wrong trig ratio:
    • using sine instead of cosine
    • using tangent instead of sine
  • Forgetting to set the calculator to degree mode.
  • Entering the ratio wrongly into the calculator for inverse trig.
  • Writing the angle of depression instead of the angle of elevation, or vice versa, without checking the diagram.
  • Forgetting that angles of elevation and depression are measured from the horizontal, not from the vertical.
  • Measuring a bearing from East or West instead of from North.
  • Measuring a bearing anticlockwise instead of clockwise.
  • Forgetting to write bearings using 3 digits, for example writing 45° instead of 045°.
  • Rounding too early in working, causing inaccurate final answers.
  • Forgetting units such as m, cm, km.
  • Not checking whether the answer is reasonable:
    • a side longer than the hypotenuse is impossible
    • an angle in a right-angled triangle must be less than 90°

Exam Tips

  • Start by drawing a clear labelled diagram if one is not given.
  • Mark the right angle clearly.
  • Label the side lengths and the angle you are using.
  • Write which trig ratio you are using:
    • “Using sinθ=OH \sin \theta = \frac{O}{H}
    • “Using tanθ=OA \tan \theta = \frac{O}{A}
  • In bearing questions, always draw a North line first.
  • State clearly:
    • “Bearing is measured clockwise from North.”
  • For angle of elevation/depression questions, include phrases such as:
    • “angle between the horizontal and line of sight”
  • Use inverse trig correctly when finding angles:
    • θ=sin1() \theta = \sin^{-1}(\dots)
    • θ=cos1() \theta = \cos^{-1}(\dots)
    • θ=tan1() \theta = \tan^{-1}(\dots)
  • Keep full calculator values until the final step, then round the final answer appropriately.
  • If the question asks for a bearing, give it in 3-digit form.
  • If the question asks for working, do not skip formula steps.
  • Always include the final statement, for example:
    • “Therefore, the height of the building is 18.4 m.”
    • “Hence, the bearing of the ship from the port is 132°.”

Quick Summary

  • Trigonometry is used to relate angles and sides in a right-angled triangle.
  • The hypotenuse is opposite the 90° angle and is the longest side.
  • Relative to a chosen angle:
    • opposite is across from the angle
    • adjacent is next to the angle, not the hypotenuse
  • Remember SOH-CAH-TOA:
    • sinθ=OH \sin \theta = \frac{O}{H}
    • cosθ=AH \cos \theta = \frac{A}{H}
    • tanθ=OA \tan \theta = \frac{O}{A}
  • To find a side, choose the trig ratio that matches the known and unknown sides.
  • To find an angle, use inverse trig:
    • sin1 \sin^{-1} , cos1 \cos^{-1} , tan1 \tan^{-1}
  • Make sure the calculator is in degree mode.
  • Angle of elevation is measured upwards from the horizontal.
  • Angle of depression is measured downwards from the horizontal.
  • A bearing is measured clockwise from North and written using 3 digits.
  • Always draw and label diagrams clearly before solving.
  • Check that answers are sensible and include correct units.
✏️ 24 practice questions available

30 questions from school exam papers

Q1

Find angle QPR.

Same diagrams as in part (a) and (b)
📊 Diagram: Same diagrams as in part (a) and (b)
2 marks
Math_Sec2NA_SA2_2023_Bedok_View_Sec 2023
Q2

Calculate ∠BAC.

Diagram for question 12b
2 marks
Math_Sec2NA_SA2_2023_Canberra_Sec 2023
Q3

Find angle QRP.

Diagram for question 5a
2 marks
Math_Sec2NA_SA2_2023_Canberra_Sec 2023
Q4

Hence, write down the smallest value of t if t is an integer.

Diagram for question 1b
2 marks
Math_Sec2NA_SA2_2023_Chung_Cheng_Yishun 2023
Q5

Find the perpendicular distance from B to AC.

Diagram for question 5c
1 mark
Math_Sec2NA_SA2_2023_Chung_Cheng_Yishun 2023
Q6

Find angle PQR.

Diagram for question 11b
2 marks
Math_Sec2NA_SA2_2023_Zhonghua_Sec 2023
Q7

The diagram shows two right angled triangles, ABD and BCD. AB = 17 cm, AD = 8 cm, BC = 12 cm, BD = 15 cm and CD = 9 cm. Find, in fractions in its simplest form, the values of (a) sin ∠BDC, and (b) tan ∠ABD.

Two right-angled triangles ABD and BCD sharing side BD. Triangle ABD has vertices A (top), B (bottom left), D (right), with AB = 17 cm, AD = 8 cm, and BD = 15 cm. Triangle BCD has vertices B (left), C (bottom right), D (top right), with BC = 12 cm, CD = 9 cm, and BD = 15 cm. Right angles are marked at A (in triangle ABD) and C (in triangle BCD).
📊 Diagram: Two right-angled triangles ABD and BCD sharing side BD. Triangle ABD has vertices A (top), B (bottom left), D (right), with AB = 17 cm, AD = 8 cm, and BD = 15 cm. Triangle BCD has vertices B (left), C (bottom right), D (top right), with BC = 12 cm, CD = 9 cm, and BD = 15 cm. Right angles are marked at A (in triangle ABD) and C (in triangle BCD).
Math_Sec2_SA2_2023_Bedok_South_Sec 2023
Q8

Show that triangle CDE is a right-angled triangle.

A diagram showing two triangles ADE and CDE. Triangle ADE has sides AE = 60 cm, DE = 100 cm, and AE is marked as 60. Triangle CDE shares side DE with triangle ADE. The measurements shown are: CD = 28 cm, CE = 96 cm, and DE = 100 cm. Point B lies on segment AC. Point D is at the left, Point E is at the right, and Point A is at the top.
📊 Diagram: A diagram showing two triangles ADE and CDE. Triangle ADE has sides AE = 60 cm, DE = 100 cm, and AE is marked as 60. Triangle CDE shares side DE with triangle ADE. The measurements shown are: CD = 28 cm, CE = 96 cm, and DE = 100 cm. Point B lies on segment AC. Point D is at the left, Point E is at the right, and Point A is at the top.
2 marks
Math_Sec2_SA2_2023_Bedok_South_Sec 2023
Q9

Find the shortest distance from C to DE.

Same diagram as question 8a showing triangles ADE and CDE with the given measurements.
📊 Diagram: Same diagram as question 8a showing triangles ADE and CDE with the given measurements.
2 marks
Math_Sec2_SA2_2023_Bedok_South_Sec 2023
Q10

Calculate ∠AED.

Same diagram as question 8a showing triangles ADE and CDE with the given measurements.
📊 Diagram: Same diagram as question 8a showing triangles ADE and CDE with the given measurements.
2 marks
Math_Sec2_SA2_2023_Bedok_South_Sec 2023
Q11

Hence, find the exact value of sin ABC.

Diagram for question 10b(i)
1 mark
Math_Sec2_SA2_2023_Broadrick_Sec 2023
Q12

Hence, find the exact value of tan ACB.

Diagram for question 10b(ii)
1 mark
Math_Sec2_SA2_2023_Broadrick_Sec 2023
Q13

x yellow balls are added to the bag. The probability of getting a yellow ball becomes 1/2. Find the value of x.

Same triangle ABC with perpendicular AD to BC at point D.
📊 Diagram: Same triangle ABC with perpendicular AD to BC at point D.
2 marks
Math_Sec2_SA2_2023_Broadrick_Sec 2023
Q14

Show that the shortest distance from D to AC is 8.42 cm.

Same triangle ABC with perpendicular AD to BC at point D.
📊 Diagram: Same triangle ABC with perpendicular AD to BC at point D.
3 marks
Math_Sec2_SA2_2023_Broadrick_Sec 2023
Q15

The total surface area of the cuboid is 325 cm². Form an equation, in terms of x, to represent this information and show that it simplifies to 18x² + 65x - 275 = 0.

Same diagram as part (a)
📊 Diagram: Same diagram as part (a)
1 mark
Math_Sec2_SA2_2023_Bukit_Merah_Sec 2023
Q16

Find angle YXZ.

Diagram for question (b)
2 marks
Math_Sec2_SA2_2023_Bukit_Merah_Sec 2023
Q17

Find the shortest distance from the point B to the line AD.

Diagram for question 9c
2 marks
Math_Sec2_SA2_2023_Bukit_Merah_Sec 2023
Q18

Expressing your answers as fractions in the simplest form, find sin ∠ADC.

Diagram for question 9d(i)
1 mark
Math_Sec2_SA2_2023_Bukit_Merah_Sec 2023
Q19

ABC is a triangle and AN is perpendicular to BC. AB = 7.2 cm, AN = 4.6 cm and angle ACN = 63°. Calculate BN.

Triangle ABC with point N on base BC. AN is drawn perpendicular to BC, creating a right angle at N. The perpendicular AN has length 4.6 cm marked. Side AB is labeled 7.2 cm. Angle ACN is marked as 63°.
📊 Diagram: Triangle ABC with point N on base BC. AN is drawn perpendicular to BC, creating a right angle at N. The perpendicular AN has length 4.6 cm marked. Side AB is labeled 7.2 cm. Angle ACN is marked as 63°.
2 marks
Math_Sec2_SA2_2023_Regent_Sec 2023
Q20

In the diagram, AP is perpendicular to BC and BQ is perpendicular to AC. QA = 36 cm, PB = 24 cm, QB = 48 cm and AB = 60 cm. (a) Giving your answer as a fraction in its simplest form, find (i) sin ∠QAB,

Triangle ABC with point P on BC and point Q on AC. AP is perpendicular to BC, and BQ is perpendicular to AC. The measurements shown are: QA = 36 cm, PB = 24 cm, QB = 48 cm, AB = 60 cm, and BC = 60 cm.
📊 Diagram: Triangle ABC with point P on BC and point Q on AC. AP is perpendicular to BC, and BQ is perpendicular to AC. The measurements shown are: QA = 36 cm, PB = 24 cm, QB = 48 cm, AB = 60 cm, and BC = 60 cm.
1 mark
Math_Sec2_SA2_2023_Regent_Sec 2023
Q21

Show that ∠FCQ ≈ 60.4°, correct to 1 decimal place.

Diagram for question 6(b)(i)
3 marks
Math_Sec2_SA2_2023_Regent_Sec 2023
Q22

State the height of the vertical tower.

A diagram showing a tower with a ball being thrown from it. The trajectory of the ball is shown as a parabolic curve above the tower. A table below shows time (t) in seconds (0, 1, 1.5, 2, 2.5, 3, 4) with corresponding heights (h) in metres (6, 18, 21, 22, 21, p, 6). The equation h = 6 + 16t - 4t² is shown above the curve.
📊 Diagram: A diagram showing a tower with a ball being thrown from it. The trajectory of the ball is shown as a parabolic curve above the tower. A table below shows time (t) in seconds (0, 1, 1.5, 2, 2.5, 3, 4) with corresponding heights (h) in metres (6, 18, 21, 22, 21, p, 6). The equation h = 6 + 16t - 4t² is shown above the curve.
1 mark
Math_Sec2_SA2_2023_Regent_Sec 2023
Q23

A ladder of length 4 m leans against a vertical wall and the bottom of the ladder is 1.5 m from the wall. The safe working angle for a ladder is between 74° and 76° to the horizontal. Is the ladder in a safe position to use? Show your working to justify your decision.

A right-angled triangle showing a ladder of length 4 m leaning against a vertical wall. The horizontal distance from the wall to the bottom of the ladder is 1.5 m. The wall is shown as a vertical line with a small square indicating the right angle where the ladder meets the wall.
📊 Diagram: A right-angled triangle showing a ladder of length 4 m leaning against a vertical wall. The horizontal distance from the wall to the bottom of the ladder is 1.5 m. The wall is shown as a vertical line with a small square indicating the right angle where the ladder meets the wall.
3 marks
Math_Sec2_SA2_2023_Zhonghua_Sec 2023
Q24

Find angle BGE.

Same diagram as part (a)
📊 Diagram: Same diagram as part (a)
3 marks
Math_Sec2NA_SA2_2023_Springfield_Sec 2023

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