MathsICSE Class 9 · Trigonometry

Trigonometric Ratios and Their Values at 0°, 30°, 45°, 60° and 90° (ICSE Class 9)

Q: Which angles are standard?

The syllabus uses 0°, 30°, 45°, 60° and 90°.

Q: Do I need a calculator?

No; exact values and given square roots are used.

Updated 17 October 2026 · Fact-checked

In a right-angled triangle, sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse and tan θ = opposite/adjacent, with cosec, sec and cot as reciprocals. Given one ratio, draw the triangle and use Pythagoras to find the rest. Know the standard values: sin 30° = 1/2, sin 45° = 1/√2, sin 60° = √3/2, tan 45° = 1, tan 30° = 1/√3, tan 60° = √3.

Opposite and adjacent depend on θ, but the hypotenuse always faces the right angle; SOH CAH TOA recalls sin, cos and tan.

Illustration drawn for QuestionBuddy.

Understand Trigonometric Ratios of an Acute Angle and Their Values at Standard Angles

The six ratios. For an acute angle $θ$ in a right-angled triangle:
$ $sin θ = \frac{opp}{hyp}, cos θ = \frac{adj}{hyp}, tan θ = \frac{opp}{adj},$ $
$ $cosec θ = \frac{1}{s i n θ}, sec θ = \frac{1}{c o s θ}, cot θ = \frac{1}{t a n θ} .$ $
Also $tan θ = \frac{s i n θ}{c o s θ}$ and $sin^{2} θ + cos^{2} θ = 1$ . 'Opposite' and 'adjacent' are relative to $θ$ ; the hypotenuse is opposite the right angle.

One ratio gives all. If $tan θ = \frac{5}{12}$ , draw a right triangle with opposite $5$ and adjacent $12$ ; the hypotenuse is $25 + 144 = 13$ . So $sin θ = \frac{5}{13}$ , $cos θ = \frac{12}{13}$ . Then $\frac{s i n θ + c o s θ}{s i n θ - c o s θ} = \frac{17/13}{- 7/13} = - \frac{17}{7}$ (the angle is acute but $sin < cos$ here, so the denominator is negative).

Standard angles.

$sin$ : $0, \frac{1}{2}, \frac{1}{2}, \frac{3}{2}, 1$ at $0^{\circ}, 3 0^{\circ}, 4 5^{\circ}, 6 0^{\circ}, 9 0^{\circ}$ .
$cos$ : $1, \frac{3}{2}, \frac{1}{2}, \frac{1}{2}, 0$ at the same angles.
$tan$ : $0, \frac{1}{3}, 1, 3$ , and not defined at $9 0^{\circ}$ .

*Memory:* $sin$ goes $\frac{0}{2}, \frac{1}{2}, \frac{2}{2}, \frac{3}{2}, \frac{4}{2}$ ; $cos$ runs in reverse.

Evaluating. $sin 6 0^{\circ} cos 3 0^{\circ} + cos 6 0^{\circ} sin 3 0^{\circ} = \frac{3}{4} + \frac{1}{4} = 1$ . $sin^{2} 3 0^{\circ} + cos^{2} 6 0^{\circ} + tan^{2} 4 5^{\circ} = \frac{1}{4} + \frac{1}{4} + 1 = \frac{3}{2}$ . Substitute exact values first, then simplify.

Applications. A ladder of length $L$ at angle $θ$ to the ground reaches height $L sin θ$ and its foot is $L cos θ$ from the wall.

Level. Class IX is a school-exam class. Standard-value evaluation is routine; given-ratio chains and surd-containing expressions are at the hardest school-exam level.

Trigonometric Ratios of an Acute Angle and Their Values at Standard Angles: formulas to remember

Ratios: $sin θ = \frac{p}{h}, cos θ = \frac{b}{h}, tan θ = \frac{p}{b}$; p perpendicular, b base, h hypotenuse.
Reciprocals: $cosec θ = \frac{h}{p}, sec θ = \frac{h}{b}, cot θ = \frac{b}{p}$; Flip.
Identity: $sin^{2} θ + cos^{2} θ = 1, tan θ = \frac{s i n θ}{c o s θ}$; Always true.
30° and 60°: $sin 3 0^{\circ} = cos 6 0^{\circ} = \frac{1}{2}, sin 6 0^{\circ} = cos 3 0^{\circ} = \frac{3}{2}$; Complementary.
Tangent values: $tan 3 0^{\circ} = \frac{1}{3}, tan 4 5^{\circ} = 1, tan 6 0^{\circ} = 3$; Learn these three.

How to solve Trigonometric Ratios of an Acute Angle and Their Values at Standard Angles questions

Draw a right triangle for the given ratio, find the third side with Pythagoras, then read off the ratios.

1Label opposite, adjacent and hypotenuse for the given angle.
2Assign lengths from the given ratio (scale if needed).
3Find the missing side by Pythagoras.
4Write the required ratios as fractions.
5For standard angles, substitute the exact values and simplify surds.
6Check with $sin^{2} θ + cos^{2} θ = 1$ .

Trigonometric Ratios of an Acute Angle and Their Values at Standard Angles in real life

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.

A ladder against a wall

Suppose a 10 m ladder leans against a wall making $6 0^{\circ}$ with the ground (take $3 = 1.732$ ).

Find the height reached and the distance of the foot from the wall.

Try it first, then show the solutionSolution

Height $= 10 sin 6 0^{\circ} = 10 \times \frac{3}{2} = 53 \approx 8.66$ m.
Foot distance $= 10 cos 6 0^{\circ} = 5$ m.

Answer: About 8.66 m high; foot 5 m from the wall.

Memory hook: Height = L sin θ, foot = L cos θ.

A ramp

Imagine a straight ramp 10 m long inclined at $3 0^{\circ}$ to the ground.

How high does it rise?

Try it first, then show the solutionSolution

Height $= 10 sin 3 0^{\circ} = 10 \times \frac{1}{2} = 5$ m.

Answer: 5 m.

Memory hook: sin 30° = 1/2.

Tricks to remember Trigonometric Ratios of an Acute Angle and Their Values at Standard Angles

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

SOH CAH TOA
Sin = Opp/Hyp, Cos = Adj/Hyp, Tan = Opp/Adj.
Use it when: Writing ratios.
√0 to √4 over 2
sin 0°, 30°, 45°, 60°, 90° = √0/2, √1/2, √2/2, √3/2, √4/2.
Use it when: Remembering the table.
Sine up, cosine down
cos runs the sin list backwards.
Use it when: cos values.

Trigonometric Ratios of an Acute Angle and Their Values at Standard Angles: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: One ratio to another · 1 mark · MCQBoard level

If $sin θ = \frac{3}{5}$ ( $θ$ acute), then $cos θ$ equals:
(a) $\frac{3}{4}$
(b) $\frac{4}{5}$
(c) $\frac{5}{4}$
(d) $\frac{5}{3}$

Try it first, then show the solutionSolution

Opposite $3$ , hypotenuse $5$ , adjacent $25 - 9 = 4$ , so $cos θ = \frac{4}{5}$ .

Answer: (b) $\frac{4}{5}$

Example 2: Ratios from sides · 3 marks · Short answerBoard level

In $△ A B C$ , right-angled at $B$ , $A B = 8$ cm and $B C = 15$ cm. Find $sin A$ , $cos A$ and $tan A$ .

Try it first, then show the solutionSolution

$A C = 64 + 225 = 17$ cm.
For angle $A$ : opposite $B C = 15$ , adjacent $A B = 8$ .
$sin A = \frac{15}{17}$ , $cos A = \frac{8}{17}$ , $tan A = \frac{15}{8}$ .

Answer: $\frac{15}{17}, \frac{8}{17}, \frac{15}{8}$

Example 3: From tan to the rest · 4 marks · Short answerHarder board level

If $tan θ = \frac{5}{12}$ , find $sin θ$ , $cos θ$ , and the value of $\frac{1 - tan ^{2} θ}{1 + tan ^{2} θ}$ . Check against $cos^{2} θ - sin^{2} θ$ .

Try it first, then show the solutionSolution

Hypotenuse $= 13$ ; $sin θ = \frac{5}{13}$ , $cos θ = \frac{12}{13}$ .
$\frac{1 - 25/144}{1 + 25/144} = \frac{119/144}{169/144} = \frac{119}{169}$ .
$cos^{2} θ - sin^{2} θ = \frac{144 - 25}{169} = \frac{119}{169}$ . ✓

Answer: $\frac{5}{13}$ , $\frac{12}{13}$ , $\frac{119}{169}$

Example 4: Standard angle expressions · 4 marks · Short answerHarder board level

Evaluate (i) $cos^{2} 4 5^{\circ} + sin^{2} 6 0^{\circ} - tan^{2} 3 0^{\circ}$ (ii) $\frac{tan 4 5 ^{\circ} + tan 3 0 ^{\circ}}{1 - tan 4 5 ^{\circ} tan 3 0 ^{\circ}}$ .

Try it first, then show the solutionSolution

(i) $(\frac{1}{2})^{2} + (\frac{3}{2})^{2} - (\frac{1}{3})^{2} = \frac{1}{2} + \frac{3}{4} - \frac{1}{3} = \frac{6 + 9 - 4}{12} = \frac{11}{12}$ .
(ii) $\frac{1 + \frac{1}{3}}{1 - \frac{1}{3}} = \frac{3 + 1}{3 - 1}$ .
Rationalise: $\frac{( 3 + 1 ) ^{2}}{3 - 1} = \frac{4 + 2 3}{2} = 2 + 3$ .

Answer: (i) $\frac{11}{12}$ (ii) $2 + 3$

Example 5: A right triangle from tan and hypotenuse · 5 marks · Long answerHOTS / competency

In $△ A B C$ right-angled at $B$ , $tan A = \frac{3}{4}$ and $A C = 20$ cm. Find $A B$ , $B C$ , the perimeter, the area and $sin C$ .

Try it first, then show the solutionSolution

$tan A = \frac{B C}{A B} = \frac{3}{4}$ : let $B C = 3 k$ , $A B = 4 k$ .
$A C = 5 k = 20$ , so $k = 4$ : $B C = 12$ , $A B = 16$ .
Perimeter $= 16 + 12 + 20 = 48$ cm; area $= \frac{1}{2} \cdot 16 \cdot 12 = 96$ cm².
$sin C = \frac{A B}{A C} = \frac{16}{20} = \frac{4}{5}$ (which equals $cos A$ ).

Answer: AB = 16 cm, BC = 12 cm, perimeter 48 cm, area 96 cm², sin C = 4/5

Where students lose marks in Trigonometric Ratios of an Acute Angle and Their Values at Standard Angles

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

Mistake: Mixing opposite and adjacent
They depend on the chosen angle.
Wrong working
$sin A = \frac{8}{17}$ with $A B = 8$ adjacent.
Right working
$sin A = \frac{15}{17}$ (opposite is $B C$ ).
Do this: Mark the angle and label the sides.
Usually costs: Whole answer.
Mistake: Wrong standard value
sin 60° is √3/2, not 1/2.
Wrong working
$sin 6 0^{\circ} = \frac{1}{2}$ .
Right working
$sin 6 0^{\circ} = \frac{3}{2}$ .
Do this: Use the √n/2 pattern.
Usually costs: Whole answer.
Mistake: Skipping Pythagoras
The third side must be found first.
Wrong working
$cos θ = \frac{5}{3}$ .
Right working
Hypotenuse is the longest: $cos θ = \frac{4}{5}$ .
Do this: Draw the triangle.
Usually costs: Whole answer.

Exam tips

Class IX is examined in school papers; show the right triangle you draw for a given ratio.
Substitute exact values before simplifying.
Check ratios of acute angles: sin and cos lie between 0 and 1.
Leave surds exact unless a decimal is asked for.

Practice questions on Trigonometric Ratios of an Acute Angle and Their Values at Standard Angles

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

Q1 · 1 mark · MCQBoard level
The value of $sin 3 0^{\circ}$ is:
(a) $\frac{3}{2}$
(b) $\frac{1}{2}$
(c) $\frac{1}{2}$
(d) $1$
Hint
Standard table.
Answer
(c) $\frac{1}{2}$
Q2 · 3 marks · Short answerHarder board level
Evaluate $4 sin^{2} 6 0^{\circ} - 3 tan^{2} 3 0^{\circ}$ and $2 sin 3 0^{\circ} cos 3 0^{\circ} tan 6 0^{\circ}$ .
Hint
Substitute exact values.
Answer
$4 \cdot \frac{3}{4} - 3 \cdot \frac{1}{3} = 3 - 1 = 2$ ; $2 \cdot \frac{1}{2} \cdot \frac{3}{2} \cdot 3 = \frac{3}{2}$ .
Q3 · 3 marks · Short answerHarder board level
If $sin θ = \frac{8}{17}$ ( $θ$ acute), find $cos θ$ , $tan θ$ and $cosec θ$ .
Hint
Adjacent = 15.
Answer
$cos θ = \frac{15}{17}$ , $tan θ = \frac{8}{15}$ , $cosec θ = \frac{17}{8}$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): $sin θ$ can be greater than 1 for an acute angle.
Reason (R): The hypotenuse is the longest side, so $\frac{opposite}{hypotenuse} < 1$ .
(a) Both (A) and (R) are true and (R) explains (A).
(b) Both are true but (R) does not explain (A).
(c) (A) is true, (R) is false.
(d) (A) is false, (R) is true.
Hint
Which side is longest?
Answer
(d)
Q5 · 4 marks · Case-basedHOTS / competency
Case study: A 10 m ladder leans against a wall at $6 0^{\circ}$ to the ground (take $3 = 1.732$ ).
(a) Find the height reached.
(b) Find the distance of the foot from the wall.
(c) If the ladder is moved to make $3 0^{\circ}$ with the ground, find the new height.
Hint
Height = 10 sin θ.
Answer
(a) 5√3 ≈ 8.66 m (b) 5 m (c) 5 m

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.Define sin θ.Show answer
Opposite divided by hypotenuse.
2.tan 45°?Show answer
1.
3.sin 60°?Show answer
√3/2.
4.Which ratio is the reciprocal of cos θ?Show answer
sec θ.

Trigonometric Ratios of an Acute Angle and Their Values at Standard Angles: frequently asked questions

Which angles are standard?

The syllabus uses 0°, 30°, 45°, 60° and 90°.

Is tan 90° defined?

No.

Do I need a calculator?

No; exact values and given square roots are used.