MathematicsClass 9ICSE

Two Sides of a Triangle Hold All the Information About Its Angle

Write the six trigonometric ratios from a right-angled triangle, rebuild all six when only one is given, prove the basic identities from the Pythagoras theorem, and evaluate expressions from a single ratio.

Why does the ratio of two sides depend only on the angle?

Draw a right-angled triangle with sides

3

cm,

4

cm and

5

cm, and look at the angle

A

facing the

3

cm side. The ratio of that opposite side to the hypotenuse is

\frac{3}{5} = 0.6

Now draw a bigger triangle with the same angles:

6

cm,

8

cm and

10

cm. The same ratio is

\frac{6}{10} = 0.6

The triangle doubled and the ratio did not move. Scale it to

30

40

50

or shrink it to

1.5

2

2.5

and you still get

0.6

, because enlarging a figure multiplies every side by the same factor, and a ratio cancels that factor away.

So the number

0.6

is not a property of any one triangle. It is a property of the angle, and it deserves a name of its own. That name is the sine of

A

, written

sin A

.

That single observation is what makes trigonometry possible. Six ratios can be formed from three sides, each depends only on the angle, and once they are tabulated you can find a side you cannot reach with a tape by measuring an angle instead.

This page covers the first part of the ICSE Class 9 Mathematics chapter on trigonometrical ratios: the six ratios, finding all of them from one, the basic identities, and evaluating expressions.

Formula

What are the six trigonometric ratios of an acute angle?

Three ratios, and their three reciprocals. For an acute angle

A

in a right-angled triangle, label the sides from

A

's point of view: the hypotenuse is opposite the right angle, the opposite side faces

A

, and the adjacent side touches

A

sin A = \frac{opposite}{hypotenuse} cos A = \frac{adjacent}{hypotenuse} tan A = \frac{opposite}{adjacent}

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

Worked example. In

△ A B C

∠ B = 9 0^{\circ}

A B = 4

cm,

B C = 3

cm and

A C = 5

cm. Write all six ratios of

∠ A

.

From

A

: the opposite side is

B C = 3

, the adjacent is

A B = 4

, the hypotenuse is

A C = 5

sin A = \frac{3}{5}, cos A = \frac{4}{5}, tan A = \frac{3}{4}

cosec A = \frac{5}{3}, sec A = \frac{5}{4}, cot A = \frac{4}{3}

Now the same triangle from the other corner. For

∠ C

, the opposite side is

A B = 4

and the adjacent is

B C = 3

sin C = \frac{4}{5}, cos C = \frac{3}{5}, tan C = \frac{4}{3}

**The words opposite and adjacent swap when the angle swaps**, while the hypotenuse never changes. That is the single most common source of error in this chapter, and the cure is to write opposite to A and adjacent to A in full, every time.

Two things worth noticing at once.

- These ratios have no units. A length divided by a length is a pure number, so

sin A = 0.6

whether the sides were measured in centimetres or in metres
- **

sin A

and

cos A

can never exceed

1

**, because the hypotenuse is the longest side of a right-angled triangle — a fact you proved in the triangle inequality chapter. So

cosec A

and

sec A

are never less than

1

, while

tan A

can be as large as you like

How do you find all six ratios when only one is given?

Treat the given ratio as two sides of a right-angled triangle, find the third by the Pythagoras theorem, and read off the rest.

Worked example 1. If

sin A = \frac{5}{13}

, find the other five ratios.

sin A

is opposite over hypotenuse, so take the opposite side as

5 k

and the hypotenuse as

13 k

. Then

adjacent = (13 k)^{2} - (5 k)^{2} = 169 k^{2} - 25 k^{2} = 144 k^{2} = 12 k

cos A = \frac{12}{13}, tan A = \frac{5}{12}, cosec A = \frac{13}{5}, sec A = \frac{13}{12}, cot A = \frac{12}{5}

**Why the letter

k

matters.** The sides could be

5

cm and

13

cm, or

10

cm and

26

cm, or anything in that proportion. Writing

5 k

and

13 k

says any triangle in this proportion, and the

k

cancels in every ratio — which is the scaling idea from the opening section, written algebraically.

Worked example 2 — given a tangent. If

15 cot A = 8

, find

sin A

and

cos A

.

First,

cot A = \frac{8}{15}

, so

tan A = \frac{15}{8}

, which means opposite

= 15 k

and adjacent

= 8 k

. Then

hypotenuse = 225 k^{2} + 64 k^{2} = 289 k^{2} = 17 k

sin A = \frac{15}{17}, cos A = \frac{8}{17}

Check with the identity:

(\frac{15}{17})^{2} + (\frac{8}{17})^{2} = \frac{225 + 64}{289} = 1

, as required.

Worked example 3 — given a secant. If

sec A = \frac{25}{7}

, find

tan A

sec A

is hypotenuse over adjacent, so hypotenuse

= 25 k

and adjacent

= 7 k

opposite = 625 k^{2} - 49 k^{2} = 576 k^{2} = 24 k ⟹ tan A = \frac{24}{7}

The triples do the work. Notice how often

(3, 4, 5)

(5, 12, 13)

(8, 15, 17)

and

(7, 24, 25)

appear — the same Pythagorean triples from the Pythagoras chapter. Questions are built from them so that the third side comes out whole, and recognising a triple saves the whole calculation.

One boundary check. A question giving

sin A = \frac{13}{5}

is impossible, since a sine cannot exceed

1

. **Test any given sine or cosine against

1

before you start** — it is the fastest way to spot a misread question.

How do you prove the basic trigonometric identities?

Write each ratio in terms of the three sides and simplify — every identity in this chapter is the Pythagoras theorem in disguise.

Call the sides of the right-angled triangle

p

(opposite

A

b

(adjacent to

A

) and

h

(hypotenuse), so that

p^{2} + b^{2} = h^{2}

.

Identity 1:

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

\frac{sin A}{cos A} = \frac{p / h}{b / h} = \frac{p}{h} \times \frac{h}{b} = \frac{p}{b} = tan A

The hypotenuse cancels, which is why the tangent needs no hypotenuse at all. The same working with the ratios inverted gives

cot A = \frac{cos A}{sin A}

.

Identity 2:

sin^{2} A + cos^{2} A = 1

sin^{2} A + cos^{2} A = \frac{p ^{2}}{h ^{2}} + \frac{b ^{2}}{h ^{2}} = \frac{p ^{2} + b ^{2}}{h ^{2}} = \frac{h ^{2}}{h ^{2}} = 1

**That is the Pythagoras theorem divided by

h^{2}

— the whole proof is one substitution.

Identity 3**:

1 + tan^{2} A = sec^{2} A

. Divide

p^{2} + b^{2} = h^{2}

b^{2}

instead:

\frac{p ^{2}}{b ^{2}} + 1 = \frac{h ^{2}}{b ^{2}} ⟹ tan^{2} A + 1 = sec^{2} A

Identity 4:

1 + cot^{2} A = cosec^{2} A

, by dividing the same equation by

p^{2}

.

So one theorem divided three different ways gives three identities. Remembering the division rather than the results means you can never mix them up — the one with

tan

and

sec

comes from dividing by the adjacent side, and

sec

is the ratio that uses the adjacent side.

**Check all four on the

3

4

5

triangle**, with

sin A = \frac{3}{5}

and

cos A = \frac{4}{5}

:

-

\frac{sin A}{cos A} = \frac{3/5}{4/5} = \frac{3}{4} = tan A

(\frac{3}{5})^{2} + (\frac{4}{5})^{2} = \frac{9 + 16}{25} = 1

1 + (\frac{3}{4})^{2} = \frac{25}{16} = (\frac{5}{4})^{2} = sec^{2} A

1 + (\frac{4}{3})^{2} = \frac{25}{9} = (\frac{5}{3})^{2} = cosec^{2} A

The notation trap.

sin^{2} A

means

(sin A)^{2}

, the whole ratio squared. It does not mean

sin (A^{2})

, and

sin^{2} A

is not

sin A^{2}

. The index sits on the ratio, not on the angle, and writing the bracket in your rough work removes all doubt.

How do you evaluate an expression when one ratio is known?

Find the two ratios the expression needs, substitute, and simplify as a single fraction. There are two routes: build the triangle, or use an identity.

Worked example 1. If

sin A = \frac{3}{5}

, evaluate

\frac{sin A + cos A}{sin A - cos A}

.

The triangle is

3

4

5

, so

cos A = \frac{4}{5}

\frac{\frac{3}{5} + \frac{4}{5}}{\frac{3}{5} - \frac{4}{5}} = \frac{\frac{7}{5}}{- \frac{1}{5}} = - 7

A negative answer is perfectly possible even though every ratio of an acute angle is positive — the subtraction in the denominator made it so.

Worked example 2. If

cos A = \frac{12}{13}

, evaluate

\frac{2 sin A - 3 cos A}{4 sin A - 9 cos A}

.

Here

sin A = \frac{5}{13}

, and since every term carries a

\frac{1}{13}

it cancels:

\frac{2 ( 5 ) - 3 ( 12 )}{4 ( 5 ) - 9 ( 12 )} = \frac{10 - 36}{20 - 108} = \frac{- 26}{- 88} = \frac{13}{44}

Cancelling the common denominator first is the whole technique for expressions of this shape. Substituting the fractions and clearing them afterwards takes four times as long.

Worked example 3 — using an identity instead. If

tan θ = \frac{3}{4}

, find

sin θ + cos θ

.

The triangle is

3

4

5

again, so

sin θ + cos θ = \frac{3}{5} + \frac{4}{5} = \frac{7}{5} = 1.4

Worked example 4. If

sec θ = \frac{13}{5}

, evaluate

\frac{1 - sin θ}{1 + cos θ}

.

From

sec θ = \frac{13}{5}

: hypotenuse

13

, adjacent

5

, so opposite

= 12

and

sin θ = \frac{12}{13}

cos θ = \frac{5}{13}

\frac{1 - \frac{12}{13}}{1 + \frac{5}{13}} = \frac{\frac{1}{13}}{\frac{18}{13}} = \frac{1}{18}

Worked example 5 — an identity does it in one line. If

sin θ = \frac{3}{5}

, evaluate

\frac{1}{sin ^{2} θ} - \frac{1}{tan ^{2} θ}

.

Rather than substituting, notice that

\frac{1}{s i n ^{2} θ} = cosec^{2} θ

and

\frac{1}{t a n ^{2} θ} = cot^{2} θ

, so by identity 4 the expression is

cosec^{2} θ - cot^{2} θ = 1

**and the value of

sin θ

was never needed at all.** That is worth watching for: an expression that reduces to an identity gives the same answer for every angle, and questions are set specifically to reward spotting it.

Exam tip

What layout keeps trigonometric ratios free of slips?

Draw the triangle and label all three sides before writing a single ratio. These questions are short, and nearly every lost mark is a mislabelled side rather than a wrong calculation.

- Mark the angle you are working with, then write opp, adj and hyp on the three sides from that angle's point of view. Re-label if the question switches to the other acute angle
- The hypotenuse is always opposite the right angle — it never depends on which acute angle you are using
- **Use

k

when a ratio is given**: opposite

= 5 k

, hypotenuse

= 13 k

. It shows the answer applies to every similar triangle and it cancels cleanly
- Recognise the triples

(3, 4, 5)

(5, 12, 13)

(8, 15, 17)

and

(7, 24, 25)

so the third side needs no calculation
- **Check a given sine or cosine against

1

. Anything larger is impossible and the question has been misread
- Cancel the common denominator first** in expressions like

\frac{2 s i n A - 3 c o s A}{4 s i n A - 9 c o s A}

- **Write

sin^{2} A

(sin A)^{2}

in rough work, so the index never migrates onto the angle
- Leave answers as exact fractions unless a decimal is asked for

The misconception to name.**

sin A

is not

sin \times A

— it is a single symbol standing for a ratio, so it cannot be cancelled or split. Writing

\frac{s i n A}{s i n B} = \frac{A}{B}

is the classic error, and one numerical test kills it:

\frac{s i n 3 0 ^{\circ}}{s i n 6 0 ^{\circ}}

is not

\frac{30}{60}

. **Treat

sin

as an instruction, never as a quantity.**

Did you know

Why can a sine never be more than one, but a tangent be anything?

The answer is a theorem you proved two chapters ago: in a right-angled triangle the hypotenuse is the longest side, because it faces the largest angle.

So in

sin A = \frac{opposite}{hypotenuse}

the numerator is always the smaller number, and the ratio can never reach

1

for an acute angle — it only gets there in the limiting case where the angle opens right out to

9 0^{\circ}

. The same argument caps the cosine. And since

cosec

and

sec

invert those, **they can never drop below

1

.

The tangent has no such ceiling**, because it compares two legs and neither has to be larger. Tilt the angle towards

9 0^{\circ}

and the opposite side grows while the adjacent side shrinks towards nothing, so the ratio climbs without limit — which is why

tan 9 0^{\circ}

is left undefined rather than given a value.

That unbounded tangent is the one you meet outside the classroom. A road sign reading gradient 1 in 12 is stating a tangent: the road rises

1

m for every

12

m measured horizontally, so

tan θ = \frac{1}{12} \approx 0.083 ⟹ θ \approx 4. 8^{\circ}

A wheelchair ramp built to a

1

12

gradient is therefore tilted by less than five degrees, which is why such a ramp has to be so long. **A rise of

1

m needs

12

m of run, and the tangent is what converts one into the other.

And the same ratio measures the sun.** When your own shadow is exactly as long as you are tall,

tan θ = 1

, so the sun stands at

4 5^{\circ}

— no instrument required. Every shadow is a tangent being displayed on the ground, which is the idea the next chapters turn into a method for measuring things that cannot be reached.

Exam relevance

How do trigonometric ratios feed into JEE-level mathematics?

This is foundation work of unusually long reach: the identities you prove here are still being used in Class 12 calculus.

Where it leads. Class 10 adds heights and distances, and Class 11 turns the ratios into trigonometric functions defined for every angle, not just acute ones — a JEE Main topic in its own right. From there they run into Complex Numbers, Inverse Trigonometric Functions, and every integral of the form

\int sin^{2} x d x

, which is solved by rewriting with the identities from this chapter.

The identity that never stops being used.

sin^{2} A + cos^{2} A = 1

is the single most reused line in the whole JEE syllabus. It is called a Pythagorean identity precisely because you derived it from the Pythagoras theorem, and its two companions,

1 + tan^{2} A = sec^{2} A

and

1 + cot^{2} A = cosec^{2} A

, appear in trigonometric substitutions in integration.

Where it appears in Physics. Resolving a vector into components is exactly this triangle: a force

F

at an angle

θ

to the horizontal has components

F cos θ

and

F sin θ

, and the reason

F^{2} = (F cos θ)^{2} + (F sin θ)^{2}

is the identity above. JEE Main and NEET Physics use it in every inclined-plane, projectile and equilibrium question.

Question types to expect. At this level: write the ratios, find all six from one, prove and apply the identities. In competitive papers: simplify a compound trigonometric expression, and identify which of four given statements is an identity. Assertion-reason items test the bounds — that a sine cannot exceed

1

and a secant cannot be less than

1

.

The single trap that costs marks. Treating

sin

as a multiplier. Cancelling it, or writing

sin (A + B) = sin A + sin B

, is the most penalised error in the whole subject, and it survives into JEE where the addition formulas are the correct route. Nothing may be cancelled out of a function's name.

Board versus competitive emphasis. ICSE asks for the labelled triangle, the stated identity and exact fractions; a competitive paper wants a simplified expression. The transferable habit is deriving an identity rather than recalling it — divide

p^{2} + b^{2} = h^{2}

by whichever square you need and the right identity appears.

Key takeaways

What should you be able to do with trigonometric ratios before Part 2?

Six ratios, one theorem, and four identities that all come out of it.

- A ratio of two sides depends only on the angle, because scaling a triangle cancels out of a ratio
- **

sin A = \frac{opp}{hyp}

cos A = \frac{adj}{hyp}

tan A = \frac{opp}{adj}

**, with

cosec

sec

and

cot

as their reciprocals
- Opposite and adjacent swap when the angle swaps; the hypotenuse never changes
- Given one ratio, write the two sides as multiples of

k

, find the third by the Pythagoras theorem, and read off the rest
- Learn the triples

(3, 4, 5)

(5, 12, 13)

(8, 15, 17)

(7, 24, 25)

— most questions are built from them
- **

tan A = \frac{s i n A}{c o s A}

and

sin^{2} A + cos^{2} A = 1

**, with

1 + tan^{2} A = sec^{2} A

and

1 + cot^{2} A = cosec^{2} A

from dividing the same theorem by

b^{2}

and

p^{2}

- **Sine and cosine never exceed

1

**; secant and cosecant are never below

1

; tangent is unbounded
- Cancel the common denominator first in expressions, and watch for ones that collapse to an identity
- **

sin

is an instruction, not a quantity** — it cannot be cancelled

The fastest self-test is example 5 above. Given

sin θ = \frac{3}{5}

, evaluate

\frac{1}{s i n ^{2} θ} - \frac{1}{t a n ^{2} θ}

— and notice whether you reached for the triangle or spotted the identity.

Why does the ratio of two sides depend only on the angle?

What are the six trigonometric ratios of an acute angle?

How do you find all six ratios when only one is given?

How do you prove the basic trigonometric identities?

How do you evaluate an expression when one ratio is known?

What layout keeps trigonometric ratios free of slips?

Why can a sine never be more than one, but a tangent be anything?

How do trigonometric ratios feed into JEE-level mathematics?

What should you be able to do with trigonometric ratios before Part 2?

Related study notes

Ready to put this into practice?