Trigonometry ICSE Class 9: Ratios, Standard Angles and Complementary Angles

What Trigonometry is about

Weightage: CISCE publishes no chapter weightage and no question-paper layout for ICSE Class 9 Mathematics. Class IX is examined by your school: one 80-mark paper of 3 hours plus 20 marks of internal assessment. Trigonometry is in the 7. Trigonometry part of the 2028 syllabus.Source: CISCE ICSE 2028 Mathematics syllabus, Class IX

Key concepts

The six trigonometric ratios

In $△ABC$ with $\angle B=90^{\circ }$, for the acute angle $A$:

• the hypotenuse is $AC$ (opposite the right angle);
• the perpendicular (opposite side) is $BC$;
• the base (adjacent side) is $AB$.

$\sin A=\frac{\text{perpendicular}}{\text{hypotenuse}}$, $\cos A=\frac{\text{base}}{\text{hypotenuse}}$, $\tan A=\frac{\text{perpendicular}}{\text{base}}$.

Reciprocals: $\text{cosec}A=\frac{1}{\sin A}$, $\sec A=\frac{1}{\cos A}$, $\cot A=\frac{1}{\tan A}$. Also $\tan A=\frac{\sin A}{\cos A}$.

For the other acute angle $C$, the perpendicular and base swap.

Worked example

In $△ABC$, $\angle B=90^{\circ }$, $AB=4$ cm and $BC=3$ cm. Write all six ratios of $\angle A$.

Try it first, then show the solutionSolution

1. $AC=\sqrt{16+9}=5$ cm.
2. $\sin A=\frac{3}{5}$, $\cos A=\frac{4}{5}$, $\tan A=\frac{3}{4}$.
3. $\text{cosec}A=\frac{5}{3}$, $\sec A=\frac{5}{4}$, $\cot A=\frac{4}{3}$.

Answer: 3/5, 4/5, 3/4, 5/3, 5/4, 4/3

Given one ratio, find the others

1. Draw a right triangle and mark the angle.
2. Label two sides from the given ratio (for example $\tan \theta =\frac{5}{12}$: perpendicular $5k$, base $12k$).
3. Find the third side by Pythagoras.
4. Read off the required ratios. The $k$ cancels.

For an acute angle all six ratios are positive; $\sin$ and $\cos$ are less than 1, while $\sec$ and $\text{cosec}$ are greater than 1.

Worked example

If $\tan \theta =\frac{5}{12}$, find $\frac{\sin \theta +\cos \theta }{\sin \theta -\cos \theta }$.

Try it first, then show the solutionSolution

1. Perpendicular $5k$, base $12k$, hypotenuse $13k$.
2. $\sin \theta =\frac{5}{13}$, $\cos \theta =\frac{12}{13}$.
3. $\frac{17/13}{-7/13}=-\frac{17}{7}$.

Answer: −17/7

Ratios of standard angles

For $0^{\circ },30^{\circ },45^{\circ },60^{\circ },90^{\circ }$ in order:

• $\sin$: $0,\frac{1}{2},\frac{1}{\sqrt{2}},\frac{\sqrt{3}}{2},1$
• $\cos$: $1,\frac{\sqrt{3}}{2},\frac{1}{\sqrt{2}},\frac{1}{2},0$
• $\tan$: $0,\frac{1}{\sqrt{3}},1,\sqrt{3}$, not defined

Where they come from: half of an equilateral triangle of side 2 has sides 1, $\sqrt{3}$, 2 and angles $30^{\circ }$, $60^{\circ }$; half of a square of side 1 has sides 1, 1, $\sqrt{2}$ and angle $45^{\circ }$.

Reciprocals follow: $\text{cosec}30^{\circ }=2$, $\sec 60^{\circ }=2$, $\cot 45^{\circ }=1$, $\sec 45^{\circ }=\sqrt{2}$, and so on. $\tan 90^{\circ }$, $\sec 90^{\circ }$, $\text{cosec}{0}^{\circ }$, $\cot 0^{\circ }$ are not defined.

Worked example

Evaluate $\sin 30^{\circ }\cos 60^{\circ }+\cos 30^{\circ }\sin 60^{\circ }$.

Try it first, then show the solutionSolution

1. $\frac{1}{2}\times \frac{1}{2}+\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}$.
2. $=\frac{1}{4}+\frac{3}{4}=1$.

Answer: 1

Finding angles from ratio values

When an equation gives a standard value, read the angle off the table. For example $2\sin \theta =\sqrt{3}$ gives $\sin \theta =\frac{\sqrt{3}}{2}$, so $\theta =60^{\circ }$ (for an acute angle).

If the angle is a multiple, solve in steps: $2\cos 2\theta =1$ gives $\cos 2\theta =\frac{1}{2}$, so $2\theta =60^{\circ }$ and $\theta =30^{\circ }$.

As the angle grows from $0^{\circ }$ to $90^{\circ }$, sine and tangent increase while cosine decreases.

Worked example

If $\sqrt{3}\tan (2\theta -15^{\circ })=3$, find the acute angle $\theta$.

Try it first, then show the solutionSolution

1. $\tan (2\theta -15^{\circ })=\frac{3}{\sqrt{3}}=\sqrt{3}=\tan 60^{\circ }$.
2. $2\theta -15^{\circ }=60^{\circ }$, so $\theta =37.5^{\circ }$.

Answer: θ = 37.5°

Complementary angles

Two angles are complementary if they add to $90^{\circ }$. In right $△ABC$ ($\angle B=90^{\circ }$), $\angle A+\angle C=90^{\circ }$, and the side opposite $A$ is adjacent to $C$. So:

$\sin A=\cos (90^{\circ }-A)$, $\cos A=\sin (90^{\circ }-A)$,
$\tan A=\cot (90^{\circ }-A)$, $\cot A=\tan (90^{\circ }-A)$,
$\sec A=\text{cosec}(90^{\circ }-A)$, $\text{cosec}A=\sec (90^{\circ }-A)$.

Use: $\frac{\sin 36^{\circ }}{\cos 54^{\circ }}=\frac{\cos 54^{\circ }}{\cos 54^{\circ }}=1$, and $\tan 10^{\circ }\tan 80^{\circ }=\tan 10^{\circ }\cot 10^{\circ }=1$.

If $\sin x=\cos y$ with both angles acute, then $x+y=90^{\circ }$.

Worked example

If $\sin 3A=\cos (A-26^{\circ })$, where $3A$ is acute, find $A$.

Try it first, then show the solutionSolution

1. $\cos (A-26^{\circ })=\sin (90^{\circ }-(A-26^{\circ }))=\sin (116^{\circ }-A)$.
2. $3A=116^{\circ }-A$, so $4A=116^{\circ }$.
3. $A=29^{\circ }$.

Answer: A = 29°

Problems with one right-angled triangle

1. Sketch the situation and mark the right angle.
2. Mark the known angle and the known side.
3. Choose the ratio that links the known side with the side you want:

• perpendicular and hypotenuse → $\sin$;
• base and hypotenuse → $\cos$;
• perpendicular and base → $\tan$.

4. Solve and give units; use $\sqrt{3}=1.732$ if a decimal is needed.

The syllabus limits these to one right-angled triangle (two-triangle heights and distances come in Class X).

Worked example

A ladder 10 m long rests against a vertical wall, making $60^{\circ }$ with the ground. How high up the wall does it reach, and how far is its foot from the wall?

Try it first, then show the solutionSolution

1. Height $=10\sin 60^{\circ }=5\sqrt{3}\approx 8.66$ m.
2. Foot distance $=10\cos 60^{\circ }=5$ m.

Answer: ≈ 8.66 m; 5 m

Maths in real life: Trigonometry

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Complementary Angles, Finding the Angle and Right-Triangle Problems
• Trigonometric Ratios of an Acute Angle and Their Values at Standard Angles

Trigonometry formula sheet

Basic ratios

$\sin A=\frac{P}{H},\cos A=\frac{B}{H},\tan A=\frac{P}{B}$

P perpendicular, B base, H hypotenuse.

Reciprocals

$\text{cosec}A=\frac{1}{\sin A},\sec A=\frac{1}{\cos A},\cot A=\frac{1}{\tan A}$

Same angle.

Quotient

$\tan A=\frac{\sin A}{\cos A}$

cot A = cos A / sin A.

30° and 60°

$\sin 30^{\circ }=\cos 60^{\circ }=\frac{1}{2};\sin 60^{\circ }=\cos 30^{\circ }=\frac{\sqrt{3}}{2}$

tan 30° = 1/√3, tan 60° = √3.

45°

$\sin 45^{\circ }=\cos 45^{\circ }=\frac{1}{\sqrt{2}};\tan 45^{\circ }=1$

Half a square.

Complementary angles

$\sin A=\cos (90^{\circ }-A);\tan A=\cot (90^{\circ }-A)$

Also sec A = cosec (90° − A).

Trigonometry: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Trigonometry

The question patterns to expect, and the method that works for each.

Tricks to remember Trigonometry

Short memory aids. Say each one aloud once, then use it on a question the same day.

• SOH CAH TOA

  Sin = Opposite/Hypotenuse, Cos = Adjacent/Hypotenuse, Tan = Opposite/Adjacent.

  Use it when: Writing the ratios.

• Root count for sine

  sin 0°, 30°, 45°, 60°, 90° = √0/2, √1/2, √2/2, √3/2, √4/2. Cosine is the same list backwards.

  Use it when: Recalling the table.

• Co- swaps with the complement

  Every ratio equals its 'co-' partner of the complementary angle: sin ↔ cos, tan ↔ cot, sec ↔ cosec.

  Use it when: Complementary angle questions.

• Angles that add to 90 cancel

  Spot pairs like 36° and 54°, 10° and 80°, 20° and 70°; they turn into the same ratio.

  Use it when: Simplifying expressions quickly.

Trigonometry: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section B style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 10-mark question split into 3 + 3 + 4 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Trigonometry

The slips examiners see most often, and what to write instead.

• ✗Mistake: Mixing up perpendicular and base

  Which side is 'opposite' depends on the angle named; students reuse the sides from the other acute angle.

  ✓Do this: Mark the angle with an arc and write P, B, H on the sketch each time.

  Usually costs: Every ratio in the question.

• ✗Mistake: Writing sin² 60° as sin 60²

  ${\sin }^{2}60^{\circ }$ means $(\sin 60^{\circ }{)}^2=\frac{3}{4}$, not $\sin 3600^{\circ }$.

  ✓Do this: Square the value: ${\left(\frac{\sqrt{3}}{2}\right)}^2=\frac{3}{4}$.

  Usually costs: 1 mark.

• ✗Mistake: Treating sin A as sin × A

  $\frac{\sin 2A}{\sin A}$ is not 2; 'sin' is not a number that cancels.

  ✓Do this: Find the value of each ratio before simplifying.

  Usually costs: The answer.

• ✗Mistake: Wrong complementary pair

  Students write $\sin A=\sin (90^{\circ }-A)$ instead of $\cos (90^{\circ }-A)$.

  ✓Do this: Each ratio swaps to its 'co-' partner: sin ↔ cos, tan ↔ cot, sec ↔ cosec.

  Usually costs: 1 to 2 marks.

• ✗Mistake: No sketch in a word problem

  Without a figure, method marks for choosing the ratio are hard to award.

  ✓Do this: Always draw and label the right triangle.

  Usually costs: 1 mark.

Practice questions on Trigonometry

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   $\sin 60^{\circ }\cos 30^{\circ }+\sin 30^{\circ }\cos 60^{\circ }$ equals:
   (a) 0
   (b) 1
   (c) $\frac{1}{2}$
   (d) $\frac{\sqrt{3}}{2}$

   Hint

   3/4 + 1/4.

   Answer

   (b)

2. Q2 · 2 marks · Short answerExam level

   Evaluate $2{\tan }^{2}45^{\circ }+{\cos }^{2}30^{\circ }-{\sin }^{2}60^{\circ }$.

   Hint

   The last two cancel.

   Answer

   $2+\frac{3}{4}-\frac{3}{4}=2$.

3. Q3 · 2 marks · Short answerExam level

   Find $\theta$ if $2\sin \theta =\sqrt{3}$ and if $2\cos 2\theta =1$ ($\theta$ acute).

   Hint

   Standard values.

   Answer

   $\theta =60^{\circ }$; $2\theta =60^{\circ }$, so $\theta =30^{\circ }$.

4. Q4 · 1 mark · Assertion-reasonHarder exam level

   Assertion (A): $\frac{\tan 35^{\circ }}{\cot 55^{\circ }}=1$.
   Reason (R): $\tan A=\cot (90^{\circ }-A)$.
   (a) (A) is true and (R) is false.
   (b) (A) is false and (R) is true.
   (c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
   (d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).

   Hint

   cot 55° = tan 35°.

   Answer

   (c)

5. Q5 · 3 marks · Short answerHarder exam level

   If $5\cos \theta =3$, evaluate $\frac{\text{cosec}\theta -\cot \theta }{\text{cosec}\theta +\cot \theta }$.

   Hint

   Triangle 3-4-5.

   Answer

   $\sin \theta =\frac{4}{5}$: $\text{cosec}\theta =\frac{5}{4}$, $\cot \theta =\frac{3}{4}$. Value $=\frac{2/4}{8/4}=\frac{1}{4}$.

6. Q6 · 3 marks · Short answerHarder exam level

   Without tables, evaluate $\frac{\sin 25^{\circ }}{\cos 65^{\circ }}+\frac{\tan 23^{\circ }}{\cot 67^{\circ }}-\frac{\sec 40^{\circ }}{\text{cosec}50^{\circ }}$.

   Hint

   Each fraction is 1.

   Answer

   $1+1-1=1$.

7. Q7 · 3 marks · Short answerHarder exam level

   In $△PQR$, $\angle Q=90^{\circ }$, $PQ=6$ cm and $\angle R=30^{\circ }$. Find $PR$ and $QR$.

   Hint

   PQ is opposite R.

   Answer

   $\sin 30^{\circ }=\frac{6}{PR}\Rightarrow PR=12$ cm; $\tan 30^{\circ }=\frac{6}{QR}\Rightarrow QR=6\sqrt{3}\approx 10.39$ cm.

8. Q8 · 3 marks · Short answerExam level

   If $A=30^{\circ }$, verify that $\sin 2A=2\sin A\cos A$ and that $\cos 2A\ne 2\cos A$.

   Hint

   Substitute.

   Answer

   $\sin 60^{\circ }=\frac{\sqrt{3}}{2}$ and $2\times \frac{1}{2}\times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}$ ✓. $\cos 60^{\circ }=\frac{1}{2}$ but $2\cos 30^{\circ }=\sqrt{3}$.

9. Q9 · 3 marks · Short answerHarder exam level

   A 15 m long ladder reaches a window when it makes $60^{\circ }$ with the ground. Find the height of the window and the distance of the ladder's foot from the wall.

   Hint

   sin and cos of 60°.

   Answer

   Height $=15\times \frac{\sqrt{3}}{2}\approx 12.99$ m; distance $=7.5$ m.

10. Q10 · 3 marks · Short answerHarder exam level

    If $\sec 4A=\text{cosec}(A-20^{\circ })$, where $4A$ is acute, find $A$.

    Hint

    sec x = cosec (90° − x).

    Answer

    $4A+A-20^{\circ }=90^{\circ }\Rightarrow A=22^{\circ }$.

11. Q11 · 4 marks · Long answerHOTS / competency

    In right $△ABC$ with $\angle B=90^{\circ }$, $\tan A=\frac{1}{\sqrt{3}}$. Find $\sin A\cos C+\cos A\sin C$ and $\cos A\cos C-\sin A\sin C$.

    Hint

    A = 30°, C = 60°.

    Answer

    $\frac{1}{2}\cdot \frac{1}{2}+\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{3}}{2}=1$; $\frac{\sqrt{3}}{2}\cdot \frac{1}{2}-\frac{1}{2}\cdot \frac{\sqrt{3}}{2}=0$.

12. Q12 · 4 marks · Long answerHOTS / competency

    In $△ABC$ right-angled at $B$, $AB+AC=9$ cm and $BC=3$ cm. Find $AB$, $AC$ and $\sin A$.

    Hint

    Let AB = x; AC = 9 − x; Pythagoras.

    Answer

    $(9-x{)}^2=x^2+9\Rightarrow 81-18x=9\Rightarrow x=4$. $AB=4$ cm, $AC=5$ cm, $\sin A=\frac{3}{5}$ (and $\cos A=\frac{4}{5}$).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Define sin A in a right triangle.Show answer

   Perpendicular (opposite side) ÷ hypotenuse.

2. 2.Value of tan 60°?Show answer

   √3.

3. 3.Value of cos 45°?Show answer

   1/√2.

4. 4.Value of cosec 30°?Show answer

   2.

5. 5.Which ratios are not defined at 90°?Show answer

   tan 90° and sec 90°.

6. 6.sin A = cos (…)?Show answer

   cos (90° − A).

7. 7.tan 25° × tan 65° = ?Show answer

   1.

8. 8.If 2 cos θ = 1 (θ acute), θ = ?Show answer

   60°.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you write all six ratios for either acute angle?
• ☐Can you find the other ratios from one ratio?
• ☐Do you know the table for 0°, 30°, 45°, 60°, 90°?
• ☐Do you square values correctly (sin² 60° = 3/4)?
• ☐Can you simplify using complementary angles?
• ☐Can you solve sin 3A = cos (A − 26°) type equations?
• ☐Do you sketch and label every word problem?

Trigonometry: frequently asked questions