Right-Angle Trigonometry · Study Notes · Edexcel IGCSE 4MA1 Higher

SOH CAH TOA: find sides and angles in right-angled triangles. Foundation for all trigonometry on Higher Tier.

At a glance

SOH: $\sin = \dfrac{O}{H}$ .
CAH: $\cos = \dfrac{A}{H}$ .
TOA: $\tan = \dfrac{O}{A}$ .
Right-angled triangles only.
Finding angle: use INVERSE trig.
Calculator in DEGREE mode.
Memorise exact values: $0°, 30°, 45°, 60°, 90°$ .

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

4.8 — Use sin, cos, tan to find sides of right-angled triangles.
4.8 — Use inverse trig to find angles.
4.8 — Solve problems involving angle of elevation/depression.
4.8 — Recall exact values of sin, cos, tan for common angles.

SOH CAH TOA

▼

Three ratios. Match the angle to its sides.

Setup. In a right-angled triangle, label sides RELATIVE to the angle $\theta$ you're interested in:

Hypotenuse: longest, opposite right angle.
Opposite: across from $\theta$ .
Adjacent: next to $\theta$ (not the hypotenuse).

The three ratios:

$\sin \theta = \dfrac{O}{H}, \quad \cos \theta = \dfrac{A}{H}, \quad \tan \theta = \dfrac{O}{A}$

Mnemonic: SOH CAH TOA.

Opposite faces θ; Adjacent touches θ; Hypotenuse faces the right angle.

Method to find a side:

Identify the angle and which sides you have/want.
Choose the right ratio.
Solve for the unknown.

Example. Hypotenuse $10$ , angle $30°$ . Find opposite.

SOH: $\sin 30° = O/10 \to O = 10 \sin 30° = 5$ .

Example. Adjacent $8$ , angle $50°$ . Find opposite.

TOA: $\tan 50° = O/8 \to O = 8 \tan 50° \approx 9.53$ .

Method to find an angle:

Use INVERSE trig functions: $\sin^{-1}$ , $\cos^{-1}$ , $\tan^{-1}$ .

Example. Opposite $7$ , adjacent $24$ .

TOA: $\tan \theta = 7/24$ .
$\theta = \tan^{-1}(7/24) \approx 16.3°$ .

Worked qualitative. Why does the ratio depend on the angle?

Right triangles with the same angles are SIMILAR.
Corresponding sides have a fixed ratio for each angle.
$\sin 30°$ is always $0.5$ , regardless of triangle size.

Edexcel tip. Always state which ratio you're using: 'using SOH', or 'tan ratio'. Mark schemes credit this.

SOH CAH TOA mnemonic.
Identify hyp, opp, adj first.
Find side: rearrange the ratio.
Find angle: use inverse trig.
Calculator in DEG mode.

Exact trig values

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Memorise these — useful for non-calculator problems and exact answers.

Memorise these exact values:

$\theta$	$\sin$	$\cos$	$\tan$
$0°$	$0$	$1$	$0$
$30°$	$\dfrac{1}{2}$	$\dfrac{\sqrt{3}}{2}$	$\dfrac{1}{\sqrt{3}}$
$45°$	$\dfrac{\sqrt{2}}{2}$	$\dfrac{\sqrt{2}}{2}$	$1$
$60°$	$\dfrac{\sqrt{3}}{2}$	$\dfrac{1}{2}$	$\sqrt{3}$
$90°$	$1$	$0$	undefined

Memory tricks:

$\sin$ : $0, 1/2, \sqrt{2}/2, \sqrt{3}/2, 1$ (over $2$ , with $\sqrt{0}, \sqrt{1}, \sqrt{2}, \sqrt{3}, \sqrt{4}$ ).
$\cos$ : same sequence in reverse.
$\tan$ : $\sin/\cos$ .

Worked qualitative. Why is $\sin 30° = 1/2$ exact?

Equilateral triangle with side $2$ . Drop perpendicular from one vertex.
Splits into two right triangles with hypotenuse $2$ , opposite $1$ (half the base).
$\sin 30° = 1/2$ — exact.

Edexcel tip. When 'exact' or 'in surd form' is asked, use these values without calculator.

Memorise common angles.
$\sin 30° = 1/2$ , $\sin 60° = \sqrt{3}/2$ .
$\sin 45° = \sqrt{2}/2$ .
Used for surd-form answers.

Angles of elevation and depression

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Looking up = elevation. Looking down = depression.

Angle of elevation: angle from HORIZONTAL looking UP.

Angle of depression: angle from HORIZONTAL looking DOWN.

Both are measured from the horizontal — NOT from the vertical.

Elevation looks up, depression looks down — both measured from the horizontal, and the two angles are equal.

Worked example. From $30$ m away, top of tree appears at angle $\theta$ above horizontal. Tree is $20$ m tall. Find $\theta$ .

Right triangle: opposite $= 20$ , adjacent $= 30$ .
$\tan \theta = 20/30 = 2/3$ .
$\theta = \tan^{-1}(2/3) \approx 33.7°$ .

From a height looking down. Same setup, just the angle is below horizontal.

Example: From a $50$ m cliff, you spot a boat. Angle of depression $20°$ . How far is the boat?

Adjacent (horizontal) is what we want. Opposite (cliff height) is $50$ .
$\tan 20° = 50/x \to x = 50/\tan 20° \approx 137$ m.

Edexcel tip. Always sketch a clear diagram. Label horizontal and vertical clearly.

Elevation: looking up from horizontal.
Depression: looking down.
Both from horizontal.
Sketch with right triangle.

Quick recap

SOH CAH TOA.
Hyp, opp, adj relative to angle.
Find side: solve for unknown.
Find angle: inverse trig.
Calculator in DEG mode.
Memorise exact values.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

SOH: sin = O/H.
CAH: cos = A/H.
TOA: tan = O/A.
$\sin 30° = 1/2$ , $\sin 60° = \sqrt{3}/2$ .
$\sin 45° = \cos 45° = \sqrt{2}/2$ .

How it’s examined

Right-angled trig every Higher Tier paper (4-7 marks). Often combined with Pythagoras or 3D. Examiner reports flag (1) wrong ratio choice, (2) calculator in radians, (3) using $\sin$ for finding angle (should be $\sin^{-1}$ ).

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/1H Jan 2024. Last reviewed 2026-05-07.

Step-by-step solutions to past-paper-style questions on right angle trigonometry, written exactly the way a tutor would explain them at the board.

1Find a side using SOH CAH TOA
Foundation• trigonometry
▼
Question
Right triangle: hypotenuse $10$ cm, angle $30°$ to a side. Find the side opposite the $30°$ angle.
Step-by-step solution
1. Step 1
  Identify: opposite (?), hypotenuse ( $10$ ), angle ( $30°$ ).
2. Step 2
  SOH: $\sin = \dfrac{\text{opp}}{\text{hyp}}$ .
3. Step 3
  $\sin 30° = \dfrac{\text{opp}}{10}$ .
4. Step 4
  $\text{opp} = 10 \sin 30° = 10 \times 0.5 = 5$ cm.
Answer
$5$ cm
2Find an angle
Higher• Adapted from 4MA1/1H Jan 2024 Q11• trigonometry, angle
▼
Question
Right triangle: opposite $7$ , adjacent $24$ . Find the angle.
Step-by-step solution
1. Step 1
  TOA: $\tan = \dfrac{\text{opp}}{\text{adj}} = \dfrac{7}{24}$ .
2. Step 2
  $\theta = \tan^{-1}(7/24) \approx 16.26°$ .
Answer
$\theta \approx 16.3°$
Examiner tip
Use INVERSE trig functions on calculator. $\tan^{-1}, \sin^{-1}, \cos^{-1}$ . Show the ratio.
3Angle of elevation
Higher• elevation
▼
Question
Tree is $20$ m tall. From a point on the ground $30$ m from the base, find the angle of elevation to the top.
Step-by-step solution
1. Step 1
  Right triangle: opposite (height) $= 20$ , adjacent (distance) $= 30$ .
2. Step 2
  $\tan \theta = 20/30 = 2/3$ .
3. Step 3
  $\theta = \tan^{-1}(2/3) \approx 33.7°$ .
Answer
$\theta \approx 33.7°$
4Exact trig values
Higher• exact values
▼
Question
Find exact value of $\sin 60° + \cos 30° + \tan 45°$ .
Step-by-step solution
1. Step 1
  $\sin 60° = \dfrac{\sqrt{3}}{2}$ , $\cos 30° = \dfrac{\sqrt{3}}{2}$ , $\tan 45° = 1$ .
2. Step 2
  Sum: $\dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{3}}{2} + 1 = \sqrt{3} + 1$ .
Answer
$\sqrt{3} + 1$
Examiner tip
Memorise exact values for $0°, 30°, 45°, 60°, 90°$ . Edexcel mark schemes give A1 for exact form.
5Angle of elevation from an angle of depression [1 mark]
Getting started• Paper 1H/2H style — 1 mark• angle of elevation, angle of depression
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Question
$T$ is the top of a vertical mast. $X$ is a point on the horizontal ground.

The angle of depression of $X$ from $T$ is $23^{\circ}$ .

Write down the angle of elevation of $T$ from $X$ . [1]
Step-by-step solution
1. Step 1
  The angle of depression is measured down from the horizontal through $T$ . The angle of elevation is measured up from the horizontal ground at $X$ . The two horizontals are parallel and $TX$ crosses both.
2. Step 2
  B1 Alternate angles are equal, so the angle of elevation is the same size.
  $23^{\circ}$
Answer
$23^{\circ}$
Examiner tip
Examiner tip: In any elevation or depression question, the angle of depression from the top equals the angle of elevation from the bottom. Put that angle inside the triangle at the point on the ground.

Common mistake: Writing $67^{\circ}$ , which is the angle between $TX$ and the mast, not the horizontal.
6Opposite side from the adjacent side [2 marks]
Getting started• Paper 1H/2H style — 2 marks• trigonometry, tangent, missing side
▼
Question
Triangle $DEF$ has a right angle at $E$ .

$DE = 7.3$ cm and angle $EDF = 56^{\circ}$ .

Calculate the length of $EF$ .

Give your answer correct to $3$ significant figures. [2]
Step-by-step solution
1. Step 1
  M1 From angle $D$ , $EF$ is opposite and $DE$ is adjacent. Opposite and adjacent means tangent.
  $\tan 56^{\circ} = \dfrac{EF}{7.3}$
2. Step 2
  A1 Multiply both sides by $7.3$ .
  $EF = 7.3 \tan 56^{\circ} = 10.822\ldots = 10.8$
Answer
$10.8$ cm
Examiner tip
Examiner tip: Sketch the triangle from the letters: the right angle at $E$ means $DF$ is the hypotenuse. The side you want and the side you have decide the ratio.

Common mistake: Using $\sin 56^{\circ}$ with $7.3$ as the hypotenuse, giving $6.05$ cm. $DE$ touches the right angle, so it cannot be the hypotenuse.
7Acute angle between the diagonals of a rectangle [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• trigonometry, rectangle, diagonals
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Question
$ABCD$ is a rectangle. The diagonals $AC$ and $BD$ meet at $M$ .

$AB = 18$ cm and $AC = 23$ cm.

Work out the size of the acute angle between the diagonals.

Give your answer correct to $1$ decimal place. [3]
Step-by-step solution
1. Step 1
  M1 Triangle $ABC$ has a right angle at $B$ . From angle $BAC$ , $AB$ is adjacent and $AC$ is the hypotenuse.
  $\cos BAC = \dfrac{18}{23} \quad \Rightarrow \quad BAC = 38.499\ldots^{\circ}$
2. Step 2
  M1 The diagonals of a rectangle are equal and bisect each other, so $MA = MB$ and angle $MBA = 38.499\ldots^{\circ}$ (base angles of an isosceles triangle are equal). Angle $AMD$ is the exterior angle of triangle $AMB$ , which equals the sum of the interior opposite angles.
  $AMD = 2 \times 38.499\ldots^{\circ}$
3. Step 3
  A1 Evaluate.
  $AMD = 76.999\ldots = 77.0^{\circ}$
Answer
$77.0^{\circ}$
Examiner tip
Examiner tip: The two angles at $M$ are $77.0^{\circ}$ and $180^{\circ} - 77.0^{\circ} = 103.0^{\circ}$ . Read the question for which one it wants.

Common mistake: Giving $38.5^{\circ}$ , the angle between a diagonal and a side. That scores M1 only.
8Two wires supporting a mast [4 marks]
Building confidence• Paper 1H/2H style — 4 marks• trigonometry, connected triangles
▼
Question
$AB$ is a vertical mast on horizontal ground, with $B$ at the foot.

A straight wire joins $A$ to the point $C$ on the ground. $AC = 18$ m and angle $ACB = 62^{\circ}$ .

A second straight wire joins $A$ to the point $D$ on the ground, on the other side of the mast. Angle $ADB = 40^{\circ}$ .

Work out the distance $BD$ .

Give your answer correct to $3$ significant figures. [4]
Step-by-step solution
1. Step 1
  M1 In triangle $ABC$ (right angle at $B$ ), $AB$ is opposite the $62^{\circ}$ angle and $AC$ is the hypotenuse.
  $AB = 18 \sin 62^{\circ}$
2. Step 2
  A1 Keep the full value.
  $AB = 15.893\ldots$
3. Step 3
  M1 In triangle $ABD$ , $AB$ is opposite the $40^{\circ}$ angle and $BD$ is adjacent.
  $\tan 40^{\circ} = \dfrac{15.893\ldots}{BD} \quad \Rightarrow \quad BD = \dfrac{15.893\ldots}{\tan 40^{\circ}}$
4. Step 4
  A1 Evaluate.
  $BD = 18.940\ldots = 18.9 \text{ m}$
Answer
$18.9$ m
Examiner tip
Examiner tip: The mast is the side both triangles share. Find it from the triangle with a known length, then use it in the other.

Common mistake: Using $\cos 62^{\circ}$ for the mast, which gives $BC = 8.45$ m (the distance along the ground), or multiplying by $\tan 40^{\circ}$ in the second triangle, giving $13.3$ m.
9An angle found through a shared side [5 marks]
Stretch• Paper 1H/2H style — 5 marks• trigonometry, connected triangles, accuracy
▼
Question
Triangle $ABC$ has a right angle at $B$ . $D$ is the point on $BC$ between $B$ and $C$ .

$AB = 10.5$ cm, angle $BAD = 24^{\circ}$ and $DC = 5.1$ cm.

Work out the size of angle $BAC$ .

Give your answer correct to $1$ decimal place. [5]
Step-by-step solution
1. Step 1
  M1 In triangle $ABD$ (right angle at $B$ ), $BD$ is opposite the $24^{\circ}$ angle and $AB$ is adjacent.
  $BD = 10.5 \tan 24^{\circ}$
2. Step 2
  A1 Keep the unrounded value.
  $BD = 4.6749\ldots$
3. Step 3
  M1 Add $DC$ to find the whole of $BC$ .
  $BC = 4.6749\ldots + 5.1 = 9.7749\ldots$
4. Step 4
  M1 In triangle $ABC$ , $BC$ is opposite angle $BAC$ and $AB$ is adjacent.
  $\tan BAC = \dfrac{9.7749\ldots}{10.5}$
5. Step 5
  A1 Evaluate.
  $BAC = 42.951\ldots = 43.0^{\circ}$
Answer
$43.0^{\circ}$
Examiner tip
Examiner tip: Angle $BAC$ is the whole angle at $A$ , made of $24^{\circ}$ and angle $DAC$ . It must be more than $24^{\circ}$ and less than $90^{\circ}$ .

Common mistake: Rounding $BD$ to $4.67$ before adding. Then $\tan^{-1}\left(\dfrac{9.77}{10.5}\right) = 42.9^{\circ}$ and the final A mark is lost. The June 2024 Paper 1H examiner report reminds candidates to keep working to a sufficient degree of accuracy that does not affect the final answer.
10Angle and diagonal of a kite [5 marks]
Stretch• Paper 1H/2H style — 5 marks• trigonometry, Pythagoras, kite
▼
Question
$ABCD$ is a kite with $AB = AD = 7$ cm and $CB = CD = 12$ cm.

The diagonal $BD = 10$ cm. The diagonals $AC$ and $BD$ meet at $E$ .

(a) Work out the size of angle $BCD$ . Give your answer correct to $1$ decimal place. [2]

(b) Work out the length of $AC$ . Give your answer correct to $3$ significant figures. [3]
Step-by-step solution
1. Step 1
  $AC$ is the line of symmetry of the kite, so it is perpendicular to $BD$ and bisects it: $BE = ED = 5$ cm, with right angles at $E$ .
2. Step 2
  M1 (a) In triangle $BCE$ , $BE$ is opposite angle $BCE$ and $CB$ is the hypotenuse.
  $\sin BCE = \dfrac{5}{12} \quad \Rightarrow \quad BCE = 24.624\ldots^{\circ}$
3. Step 3
  A1 Angle $BCD$ is twice angle $BCE$ .
  $BCD = 49.248\ldots = 49.2^{\circ}$
4. Step 4
  M1 (b) Pythagoras in triangle $ABE$ , where $AB$ is the hypotenuse, so subtract.
  $AE = \sqrt{7^{2} - 5^{2}} = \sqrt{24} = 4.8989\ldots$
5. Step 5
  M1 Pythagoras in triangle $CBE$ .
  $CE = \sqrt{12^{2} - 5^{2}} = \sqrt{119} = 10.908\ldots$
6. Step 6
  A1 Add the two parts.
  $AC = 4.8989\ldots + 10.908\ldots = 15.807\ldots = 15.8$
Answer
(a) $49.2^{\circ}$ (b) $15.8$ cm
Examiner tip
Examiner tip: Part (b) can also be done with trigonometry, $CE = 12 \cos 24.624\ldots^{\circ}$ , but Pythagoras avoids carrying an angle. Either earns the marks.

Common mistake: Adding the squares, $\sqrt{7^{2} + 5^{2}}$ , when the side wanted is shorter than the hypotenuse. The June 2024 Paper 1H examiner report records that adding the squares in this situation scored zero.

Question 1
Paper 1H/2H style — 1 mark1 mark
$x$ is an acute angle and $\cos x^{\circ} = \sin 25^{\circ}$

Write down the value of $x$ . [1]
Model answer
In a right-angled triangle with an angle of $25^{\circ}$ , the other acute angle is $90 - 25 = 65^{\circ}$ . The side opposite the $25^{\circ}$ angle is adjacent to the $65^{\circ}$ angle, so $\sin 25^{\circ} = \cos 65^{\circ}$ .

Answer: $x = 65$
Why this scores
Mark scheme:
- B1 for $65$
Why this scores full marks: The script uses one right-angled triangle for both ratios, so the answer comes from the angle sum of a triangle, not from a calculator search. A quick check: $\cos 65^{\circ} = 0.4226\ldots = \sin 25^{\circ}$ .
Question 2
Paper 1H/2H style — 2 marks2 marks
Triangle $RST$ has a right angle at $S$ .

$ST = 6.4$ cm and angle $SRT = 18^{\circ}$ .

Calculate the length of $RS$ .

Give your answer correct to $3$ significant figures. [2]
Model answer
From angle $R$ : $ST$ is opposite, $RS$ is adjacent.

$\tan 18^{\circ} = \dfrac{6.4}{RS}$

$RS = \dfrac{6.4}{\tan 18^{\circ}} = 19.697\ldots$

Answer: $19.7$ cm
Why this scores
Mark scheme:
- M1 for $\dfrac{6.4}{\tan 18^{\circ}}$ oe
- A1 for awrt $19.7$
Why this scores full marks: The unknown is adjacent, on the bottom of the tangent ratio, so the script divides. $6.4 \tan 18^{\circ} = 2.08$ would make the side next to a small $18^{\circ}$ angle shorter than the side opposite it, which is impossible.
Question 3
Paper 1H/2H style — 2 marks2 marks
A straight road rises $1$ m vertically for every $8$ m it goes horizontally.

Work out the angle the road makes with the horizontal.

Give your answer correct to $1$ decimal place. [2]
Model answer
The rise of $1$ m is opposite the angle and the $8$ m horizontal is adjacent.

$\tan \theta = \dfrac{1}{8}$

$\theta = \tan^{-1}(0.125) = 7.125\ldots$

Answer: $7.1^{\circ}$
Why this scores
Mark scheme:
- M1 for $\tan \theta = \dfrac{1}{8}$ oe
- A1 for awrt $7.1$
Why this scores full marks: The $8$ m is measured horizontally, so it is the adjacent side, not the length of the road. Scripts that treat $8$ m as the hypotenuse use $\sin^{-1}\left(\dfrac{1}{8}\right) = 7.18\ldots$ , which rounds to $7.2^{\circ}$ and loses the A mark.
Question 4
Paper 1H/2H style — 3 marks3 marks
A straight playground slide makes an angle of $34^{\circ}$ with the horizontal ground.

The bottom of the slide is $2.9$ m horizontally from the point on the ground directly below the top of the slide.

(a) Work out the length of the slide. Give your answer correct to $3$ significant figures. [2]

(b) Work out the height of the top of the slide above the ground. Give your answer correct to $3$ significant figures. [1]
Model answer
(a) The $2.9$ m is adjacent to the $34^{\circ}$ and the slide is the hypotenuse.

$\cos 34^{\circ} = \dfrac{2.9}{L} \quad \Rightarrow \quad L = \dfrac{2.9}{\cos 34^{\circ}} = 3.4980\ldots$

Answer: $3.50$ m

(b) The height is opposite the $34^{\circ}$ .

$h = 2.9 \tan 34^{\circ} = 1.9560\ldots$

Answer: $1.96$ m
Why this scores
Mark scheme:
- M1 for $\dfrac{2.9}{\cos 34^{\circ}}$ oe
- A1 for awrt $3.50$
- B1 for awrt $1.96$
Why this scores full marks: The unknown length is on the bottom of the cosine ratio, so the script divides; $2.9 \cos 34^{\circ} = 2.40$ would make the slide shorter than the ground it covers. The answer $3.50$ keeps the final zero, which is part of giving $3$ significant figures. Part (b) uses the given $2.9$ m with tangent rather than a rounded answer from (a).
Question 5
Paper 1H/2H style — 3 marks3 marks
Triangle $PQR$ is isosceles with $PQ = PR = 9.5$ cm.

Angle $QPR = 38^{\circ}$ .

Work out the length of $QR$ .

Give your answer correct to $3$ significant figures. [3]
Model answer
The perpendicular from $P$ to $QR$ meets it at $N$ . It bisects angle $QPR$ and the base, so angle $QPN = 19^{\circ}$ and $QR = 2 \times QN$ .

$QN = 9.5 \sin 19^{\circ} = 3.0928\ldots$

$QR = 2 \times 3.0928\ldots = 6.1857\ldots$

Answer: $6.19$ cm
Why this scores
Mark scheme:
- M1 for $19^{\circ}$ used in a right-angled triangle with hypotenuse $9.5$
- M1 for $9.5 \sin 19^{\circ}$ oe
- A1 for awrt $6.19$
Why this scores full marks: Both the angle and the base are halved by the line of symmetry, and the script doubles back at the end. The usual losses are using $\sin 38^{\circ}$ (the full angle) or forgetting to double, giving $3.09$ cm.
Question 6
Paper 1H/2H style — 3 marks3 marks
$ABCD$ is a rectangle with $AB = 16$ cm.

The diagonal $AC$ makes an angle of $33^{\circ}$ with $AB$ .

Work out the area of the rectangle.

Give your answer correct to $3$ significant figures. [3]
Model answer
In triangle $ABC$ (right angle at $B$ ), $BC$ is opposite the $33^{\circ}$ angle and $AB$ is adjacent.

$BC = 16 \tan 33^{\circ} = 10.390\ldots$

$\text{area} = 16 \times 10.390\ldots = 166.24\ldots$

Answer: $166$ cm $^2$
Why this scores
Mark scheme:
- M1 for $16 \tan 33^{\circ}$ oe
- M1 for $16 \times$ their $BC$
- A1 for awrt $166$
Why this scores full marks: The diagonal splits the rectangle into two right-angled triangles, and the script names the one it uses. Using $16 \sin 33^{\circ}$ treats $AB$ as the hypotenuse and gives $139$ cm $^2$ .
Question 7
Paper 1H/2H style — 4 marks4 marks
$ABCD$ is a parallelogram with $AB = 12$ cm, $AD = 7$ cm and angle $DAB = 58^{\circ}$ .

(a) Work out the perpendicular distance between the sides $AB$ and $DC$ . Give your answer correct to $3$ significant figures. [2]

(b) Work out the area of the parallelogram. Give your answer correct to $3$ significant figures. [2]
Model answer
(a) Drop a perpendicular from $D$ to $AB$ . $AD = 7$ is the hypotenuse and the height is opposite the $58^{\circ}$ .

$h = 7 \sin 58^{\circ} = 5.9363\ldots$

Answer: $5.94$ cm

(b) $\text{area} = 12 \times 5.9363\ldots = 71.236\ldots$

Answer: $71.2$ cm $^2$
Why this scores
Mark scheme:
- M1 for $7 \sin 58^{\circ}$
- A1 for awrt $5.94$
- M1 for $12 \times$ their height
- A1 for awrt $71.2$
Why this scores full marks: Part (b) uses the unrounded height from (a). Using the rounded $5.94$ gives $12 \times 5.94 = 71.28$ , which rounds to $71.3$ and loses the final A mark. Using the slant side, $12 \times 7 = 84$ , is the other common error: the area needs the perpendicular height.
Question 8
Paper 1H/2H style — 4 marks4 marks
Triangle $ABC$ has a right angle at $B$ . $D$ is the point on $BC$ between $B$ and $C$ .

$AB = 7$ cm, $BD = 3$ cm and $DC = 5$ cm.

Work out the size of angle $DAC$ .

Give your answer correct to $1$ decimal place. [4]
Model answer
$BC = 3 + 5 = 8$ cm.

$BAC = \tan^{-1}\left(\dfrac{8}{7}\right) = 48.814\ldots$

$BAD = \tan^{-1}\left(\dfrac{3}{7}\right) = 23.198\ldots$

$DAC = 48.814\ldots - 23.198\ldots = 25.615\ldots$

Answer: $25.6^{\circ}$
Why this scores
Mark scheme:
- M1 for $\tan^{-1}\left(\dfrac{8}{7}\right)$ or $48.8\ldots$
- M1 for $\tan^{-1}\left(\dfrac{3}{7}\right)$ or $23.2\ldots$
- M1 for subtracting their angles
- A1 for awrt $25.6$
Why this scores full marks: Triangle $ADC$ has no right angle, so the script works with the two right-angled triangles that share the vertex $A$ and subtracts. The weaker script uses $\tan^{-1}\left(\dfrac{5}{7}\right) = 35.5^{\circ}$ , treating $DC$ as if it were opposite a right-angled corner.
Question 9
Paper 1H/2H style — 5 marks5 marks
A window $W$ is $14$ m above horizontal ground. Directly across the street is a vertical building $FT$ , with foot $F$ on the ground and top $T$ .

From $W$ , the angle of depression of $F$ is $22^{\circ}$ and the angle of elevation of $T$ is $35^{\circ}$ .

Work out the height of the building $FT$ .

Give your answer correct to $3$ significant figures. [5]
Model answer
Let $d$ be the horizontal distance from $W$ to the building. The horizontal line from $W$ meets $FT$ at $G$ , $14$ m above $F$ .

Angle of depression $22^{\circ}$ , opposite side $14$ :

$d = \dfrac{14}{\tan 22^{\circ}} = 34.651\ldots$

Angle of elevation $35^{\circ}$ :

$GT = 34.651\ldots \times \tan 35^{\circ} = 24.263\ldots$

$FT = 14 + 24.263\ldots = 38.263\ldots$

Answer: $38.3$ m
Why this scores
Mark scheme:
- M1 for $\dfrac{14}{\tan 22^{\circ}}$ oe
- A1 for $34.65\ldots$
- M1 for their $d \times \tan 35^{\circ}$
- M1 for adding $14$ to their $GT$
- A1 for awrt $38.3$
Why this scores full marks: The horizontal line from the window splits the building into two parts, one in each right-angled triangle, and the script names the point $G$ so each triangle is clear. The horizontal distance links them and is kept unrounded. Weaker scripts give $24.3$ m, forgetting the $14$ m below the window.
Question 10
Paper 1H/2H style — 5 marks5 marks
A walker follows a straight path $350$ m long that rises at $8^{\circ}$ to the horizontal. She then follows a second straight path $420$ m long that rises at $14^{\circ}$ to the horizontal. Both paths head in the same direction.

Work out the angle to the horizontal of the straight line from her starting point to her finishing point.

Give your answer correct to $1$ decimal place. [5]
Model answer
Vertical rise:

$350 \sin 8^{\circ} + 420 \sin 14^{\circ} = 48.710\ldots + 101.607\ldots = 150.317\ldots$

Horizontal distance:

$350 \cos 8^{\circ} + 420 \cos 14^{\circ} = 346.593\ldots + 407.524\ldots = 754.118\ldots$

$\tan \theta = \dfrac{150.317\ldots}{754.118\ldots}$

$\theta = 11.272\ldots$

Answer: $11.3^{\circ}$
Why this scores
Mark scheme:
- M1 for $350 \sin 8^{\circ}$ or $420 \sin 14^{\circ}$
- M1 for $350 \cos 8^{\circ}$ or $420 \cos 14^{\circ}$
- M1 for both totals, $150.3\ldots$ and $754.1\ldots$
- M1 for $\tan^{-1}\left(\dfrac{\text{their rise}}{\text{their horizontal}}\right)$
- A1 for awrt $11.3$
Why this scores full marks: Angles do not add or average along a path, but vertical and horizontal distances do. The script splits each path into its two parts, adds like with like, then makes one right-angled triangle. Averaging the angles, $\dfrac{8 + 14}{2} = 11^{\circ}$ , scores nothing.

4MA1 Right-Angle Trigonometry

At a glance

What you’ll learn

SOH CAH TOA

Exact trig values

Angles of elevation and depression

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Find a side using SOH CAH TOA

2Find an angle

3Angle of elevation

4Exact trig values

5Angle of elevation from an angle of depression [1 mark]

6Opposite side from the adjacent side [2 marks]

7Acute angle between the diagonals of a rectangle [3 marks]

8Two wires supporting a mast [4 marks]

9An angle found through a shared side [5 marks]

10Angle and diagonal of a kite [5 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10