MathsICSE Class 9 · Trigonometry

Ratios of Complementary Angles, Finding the Angle and One-Right-Triangle Problems (ICSE Class 9)

Q: Do these results hold for obtuse angles?

Class IX uses them for acute angles only.

Q: Is there a calculator in this chapter?

No; exact standard values are used.

Updated 17 October 2026 · Fact-checked

For an acute angle θ: sin(90° − θ) = cos θ, cos(90° − θ) = sin θ, tan(90° − θ) = cot θ, and similarly cosec/sec. Convert each ratio to the complementary angle to evaluate products and quotients. To find an angle, match the ratio value to a standard angle; for sin A = cos B form, use A + B = 90°.

The side opposite θ is adjacent to 90° − θ, so each ratio of 90° − θ equals the co-ratio of θ.

Illustration drawn for QuestionBuddy.

Understand Complementary Angles, Finding the Angle and Right-Triangle Problems

**Complementary ratios (acute $θ$ ).**
$ $sin (9 0^{\circ} - θ) = cos θ, cos (9 0^{\circ} - θ) = sin θ, tan (9 0^{\circ} - θ) = cot θ,$ $
$ $sec (9 0^{\circ} - θ) = cosec θ, cosec (9 0^{\circ} - θ) = sec θ, cot (9 0^{\circ} - θ) = tan θ .$ $
Reason: in a right triangle the two acute angles add to $9 0^{\circ}$ ; the side opposite one is adjacent to the other.

Evaluating. $sin 3 5^{\circ} = cos 5 5^{\circ}$ , so $\frac{s i n 2 5 ^{\circ}}{c o s 6 5 ^{\circ}} + \frac{c o s 4 0 ^{\circ}}{s i n 5 0 ^{\circ}} = 1 + 1 = 2$ . For products, pair angles that add to $9 0^{\circ}$ : $tan 1 5^{\circ} tan 2 5^{\circ} tan 4 5^{\circ} tan 6 5^{\circ} tan 7 5^{\circ} = (tan 1 5^{\circ} tan 7 5^{\circ}) (tan 2 5^{\circ} tan 6 5^{\circ}) tan 4 5^{\circ}$ and $tan 7 5^{\circ} = cot 1 5^{\circ}$ , so each pair is $1$ : the value is $1$ .

Finding an angle.

Match with a standard value: $2 sin θ - 1 = 0 \Rightarrow sin θ = \frac{1}{2} \Rightarrow θ = 3 0^{\circ}$ ; $tan θ = 3 \Rightarrow 6 0^{\circ}$ ; $2 cos^{2} θ = 1 \Rightarrow cos θ = \frac{1}{2} \Rightarrow 4 5^{\circ}$ .
If $sin A = cos B$ (both acute), then $A + B = 9 0^{\circ}$ . For $sin (2 x + 10)^{\circ} = cos (x + 20)^{\circ}$ : $2 x + 10 + x + 20 = 90$ , so $x = 20$ (check $sin 5 0^{\circ} = cos 4 0^{\circ}$ ). For $sin 3 A = cos (A - 2 6^{\circ})$ with $A$ acute: $3 A + A - 26 = 90$ , so $A = 2 9^{\circ}$ .

Right-triangle problems. Draw the triangle, label the angle, and use
$ $height = distance \times tan θ, shadow = \frac{height}{t a n θ} .$ $
A $6$ m pole with sun elevation $3 0^{\circ}$ has shadow $\frac{6}{t a n 3 0 ^{\circ}} = 63 \approx 10.39$ m; at $4 5^{\circ}$ the shadow is $6$ m; at $6 0^{\circ}$ it is $\frac{6}{3} = 23 \approx 3.46$ m.

Level. Class IX is a school-exam class. Direct complementary conversions are routine; the equation $sin 3 A = cos (A - 2 6^{\circ})$ and product evaluations are at the hardest school-exam level.

Complementary Angles, Finding the Angle and Right-Triangle Problems: formulas to remember

Sine-cosine pair: $sin (9 0^{\circ} - θ) = cos θ$; And cos(90° − θ) = sin θ.
Tangent-cotangent pair: $tan (9 0^{\circ} - θ) = cot θ$; Product tan θ tan(90° − θ) = 1.
If sin A = cos B: $A + B = 9 0^{\circ}$; Both acute.
Height and shadow: $h = d tan θ$; d horizontal distance.

How to solve Complementary Angles, Finding the Angle and Right-Triangle Problems questions

Convert to complementary angles or match standard values, then solve; for diagrams use the right triangle.

1Check that the angles are acute and that pairs add up to $9 0^{\circ}$ .
2Replace $sin$ by $cos$ of the complement (and similarly for the other pairs).
3Cancel or combine; use $tan θ tan (9 0^{\circ} - θ) = 1$ .
4For an equation $sin A = cos B$ , set $A + B = 9 0^{\circ}$ and solve.
5For a diagram, mark the angle, choose the ratio that links the known and unknown sides, then solve.
6Check by evaluating both sides at the answer.

Complementary Angles, Finding the Angle and Right-Triangle Problems 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.

Shadow of a pole

Suppose a 6 m pole casts a shadow when the sun is at different elevations (take $3 = 1.732$ ).

Find the shadow lengths at $4 5^{\circ}$ and $6 0^{\circ}$ , and the sun elevation if the shadow is $63$ m.

Try it first, then show the solutionSolution

$4 5^{\circ}$ : shadow $= \frac{6}{1} = 6$ m.
$6 0^{\circ}$ : shadow $= \frac{6}{3} = 23 \approx 3.46$ m.
Shadow $63$ : $tan θ = \frac{6}{6 3} = \frac{1}{3}$ , so $θ = 3 0^{\circ}$ .

Answer: 6 m; about 3.46 m; 30°.

Memory hook: tan θ = height / shadow.

Two paired angles

Imagine a design sheet listing $sin 7 0^{\circ}$ and $cos 2 0^{\circ}$ .

How are they related?

Try it first, then show the solutionSolution

$7 0^{\circ} + 2 0^{\circ} = 9 0^{\circ}$ , so $sin 7 0^{\circ} = cos 2 0^{\circ}$ .
Their ratio is $1$ .

Answer: They are equal.

Memory hook: Complementary angles swap sine and cosine.

Tricks to remember Complementary Angles, Finding the Angle and Right-Triangle Problems

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

Co-function
Sin and cos of complementary angles swap; so do tan/cot and sec/cosec.
Use it when: Every conversion.
Pairs make 90
In a product, pair angles that add to 90°.
Use it when: tan 15° tan 75° = 1.
sin A = cos B ⇒ A + B = 90°
Add the angles and equate to 90°.
Use it when: Finding unknown angles.

Complementary Angles, Finding the Angle and Right-Triangle Problems: solved examples

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

Example 1: Complementary pair · 1 mark · MCQBoard level

$sin 3 5^{\circ}$ is equal to:
(a) $sin 5 5^{\circ}$
(b) $cos 3 5^{\circ}$
(c) $cos 5 5^{\circ}$
(d) $tan 5 5^{\circ}$

Try it first, then show the solutionSolution

$sin 3 5^{\circ} = cos (9 0^{\circ} - 3 5^{\circ}) = cos 5 5^{\circ}$ .

Answer: (c)

Example 2: Evaluating · 3 marks · Short answerBoard level

Evaluate $\frac{sin 2 5 ^{\circ}}{cos 6 5 ^{\circ}} + \frac{cos 4 0 ^{\circ}}{sin 5 0 ^{\circ}}$ .

Try it first, then show the solutionSolution

$cos 6 5^{\circ} = sin 2 5^{\circ}$ , so the first term is $1$ .
$sin 5 0^{\circ} = cos 4 0^{\circ}$ , so the second term is $1$ .
Sum $= 2$ .

Answer: 2

Example 3: A product of tangents · 3 marks · Short answerHarder board level

Evaluate $tan 1 5^{\circ} tan 2 5^{\circ} tan 4 5^{\circ} tan 6 5^{\circ} tan 7 5^{\circ}$ .

Try it first, then show the solutionSolution

$tan 7 5^{\circ} = cot 1 5^{\circ}$ , so $tan 1 5^{\circ} tan 7 5^{\circ} = 1$ .
$tan 6 5^{\circ} = cot 2 5^{\circ}$ , so $tan 2 5^{\circ} tan 6 5^{\circ} = 1$ .
$tan 4 5^{\circ} = 1$ . The value is $1$ .

Answer: 1

Example 4: Finding angles · 3 marks · Short answerHarder board level

Find the acute angle $θ$ if (i) $2 sin θ - 1 = 0$ (ii) $tan θ = 3$ (iii) $2 cos^{2} θ = 1$ .

Try it first, then show the solutionSolution

(i) $sin θ = \frac{1}{2}$ , so $θ = 3 0^{\circ}$ .
(ii) $θ = 6 0^{\circ}$ .
(iii) $cos^{2} θ = \frac{1}{2}$ , $cos θ = \frac{1}{2}$ , so $θ = 4 5^{\circ}$ .

Answer: $3 0^{\circ}$ , $6 0^{\circ}$ , $4 5^{\circ}$

Example 5: Sine equals cosine · 5 marks · Long answerHOTS / competency

Find $x$ if $sin (2 x + 10)^{\circ} = cos (x + 20)^{\circ}$ , and find the acute angle $A$ if $sin 3 A = cos (A - 2 6^{\circ})$ . Verify each.

Try it first, then show the solutionSolution

$sin (2 x + 10) = cos (x + 20) = sin (90 - x - 20)$ , so $2 x + 10 = 70 - x$ and $x = 20$ .
Check: $sin 5 0^{\circ} = cos 4 0^{\circ}$ ✓.
$sin 3 A = cos (A - 26) \Rightarrow 3 A + (A - 26) = 90$ , so $4 A = 116$ and $A = 2 9^{\circ}$ .
Check: $sin 8 7^{\circ} = cos 3^{\circ}$ ✓ (and $A - 26 = 3$ , acute).

Answer: x = 20; A = 29°

Where students lose marks in Complementary Angles, Finding the Angle and Right-Triangle Problems

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

Mistake: Applying co-function rules to non-complementary angles
The angles must add up to 90°.
Wrong working
$sin 4 0^{\circ} = cos 4 0^{\circ}$ .
Right working
$sin 4 0^{\circ} = cos 5 0^{\circ}$ .
Do this: Subtract from 90°.
Usually costs: Whole answer.
Mistake: Forgetting tan 45° = 1
The middle factor is 1, not 0.
Wrong working
Product $= 0$ .
Right working
Product $= 1$ .
Do this: Remember the exact value.
Usually costs: Final answer.
Mistake: Not checking the angle is acute
Solutions must be acute in these questions.
Wrong working
$A - 26 < 0$ accepted.
Right working
Check: $A - 26 = 3$ , acute.
Do this: Substitute back.
Usually costs: Verification mark.

Exam tips

Class IX is examined in school papers; state "since 90° − θ" whenever you change the ratio.
Pair angles in products before computing anything.
Verify equations by evaluating both sides.
In diagrams, label the angle and the sides first.

Practice questions on Complementary Angles, Finding the Angle and Right-Triangle Problems

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
$cosec 2 5^{\circ}$ is equal to:
(a) $sec 6 5^{\circ}$
(b) $sec 2 5^{\circ}$
(c) $cot 6 5^{\circ}$
(d) $tan 6 5^{\circ}$
Hint
$25 + 65 = 90$ .
Answer
(a) sec 65°
Q2 · 4 marks · Short answerHarder board level
Show that $cosec 3 1^{\circ} - sec 5 9^{\circ} = 0$ and that $tan 5^{\circ} tan 2 5^{\circ} tan 4 5^{\circ} tan 6 5^{\circ} tan 8 5^{\circ} = 1$ .
Hint
Complementary pairs.
Answer
$sec 5 9^{\circ} = cosec (90 - 59)^{\circ} = cosec 3 1^{\circ}$ , so the difference is $0$ . Pairs $(5, 85)$ , $(25, 65)$ give $1 \cdot 1$ and $tan 4 5^{\circ} = 1$ , so the product is $1$ .
Q3 · 3 marks · Short answerHarder board level
Find the acute angle $θ$ if $2 cos 3 θ = 1$ .
Hint
$cos 3 θ = \frac{1}{2}$ .
Answer
$3 θ = 6 0^{\circ}$ , so $θ = 2 0^{\circ}$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): $tan 4 5^{\circ} = cot 4 5^{\circ}$ .
Reason (R): $tan (9 0^{\circ} - θ) = cot θ$ .
(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
Put $θ = 4 5^{\circ}$ .
Answer
(a)
Q5 · 4 marks · Case-basedHOTS / competency
Case study: A pole is $6$ m tall. The sun's elevation changes through the day (take $3 = 1.732$ ).
(a) Find $tan θ$ when the shadow is $63$ m long.
(b) Find $θ$ .
(c) Find the shadow when the elevation is $4 5^{\circ}$ .
(d) Find the shadow when the elevation is $6 0^{\circ}$ .
Hint
tan θ = height ÷ shadow.
Answer
(a) $\frac{1}{3}$ (b) $3 0^{\circ}$ (c) $6$ m (d) $23 \approx 3.46$ 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.sin(90° − θ)?Show answer
cos θ.
2.tan θ × tan(90° − θ)?Show answer
1.
3.If sin A = cos B, what is A + B?Show answer
90°.
4.Shadow length for a pole of height h at elevation θ?Show answer
h divided by tan θ.

Complementary Angles, Finding the Angle and Right-Triangle Problems: frequently asked questions

Do these results hold for obtuse angles?

Class IX uses them for acute angles only.

Is there a calculator in this chapter?

No; exact standard values are used.

What are complementary angles?

Two angles that add to 90°.