MathsCBSE Class 10 · Introduction to Trigonometry

Trigonometric Ratios of an Acute Angle in a Right Triangle

Updated 13 October 2026 · Fact-checked

Trigonometric ratios of an acute angle compare pairs of sides of a right-angled triangle: sine, cosine, tangent and their reciprocals cosecant, secant and cotangent. They depend only on the angle, not on the triangle's size. The board asks you to label sides correctly, find a missing side by Pythagoras and write the ratios.

Understand Trigonometric Ratios of an Acute Angle

Take a right-angled triangle $A B C$ with $∠ B = 9 0^{\circ}$ . Fix your attention on the acute angle $A$ . Relative to $A$ , the side $B C$ across from it is the opposite side, the side $A B$ touching it (other than the longest side) is the adjacent side, and $A C$ , across from the right angle, is the hypotenuse.

A trigonometric ratio is the ratio of two of these sides. There are six: $sin A = \frac{B C}{A C}$ , $cos A = \frac{A B}{A C}$ , $tan A = \frac{B C}{A B}$ , and their reciprocals $cosec A = \frac{A C}{B C}$ , $sec A = \frac{A C}{A B}$ , $cot A = \frac{A B}{B C}$ . Each name is one symbol: $sin A$ means the sine of angle $A$ , not $sin \times A$ .

The labels move with the angle. If you stand at $∠ C$ instead, the opposite side becomes $A B$ and the adjacent side becomes $B C$ , while the hypotenuse stays $A C$ . So always mark the angle first, then name the sides. Most lost marks in this topic come from skipping this step.

Why is $sin A$ a fixed number? Draw any other right triangle $A^{'} B^{'} C^{'}$ with $∠ B^{'} = 9 0^{\circ}$ and $∠ A^{'} = ∠ A$ . Two angles match, so $△ A B C \sim △ A^{'} B^{'} C^{'}$ by AA similarity. Corresponding sides are proportional, so $\frac{B C}{A C} = \frac{B ^{'} C ^{'}}{A ^{'} C ^{'}}$ . The same holds for every ratio. So each ratio depends only on the angle: the ratios are well defined. This is also why, if $sin A = \frac{3}{5}$ , you write the sides as $3 k$ and $5 k$ , not exactly 3 and 5.

Two facts hold for every acute angle. The hypotenuse is the longest side, so $0 < sin A < 1$ and $0 < cos A < 1$ , while $sec A > 1$ and $cosec A > 1$ . And $tan A$ can be any positive number. When a side is missing, Pythagoras, $A C^{2} = A B^{2} + B C^{2}$ , gives it.

Everything else in the chapter grows from these definitions: finding all ratios from one, the values at $3 0^{\circ}$ , $4 5^{\circ}$ , $6 0^{\circ}$ , the cases $0^{\circ}$ and $9 0^{\circ}$ , and the identity $sin^{2} A + cos^{2} A = 1$ , which is just Pythagoras divided by $A C^{2}$ .

Trigonometric Ratios of an Acute Angle: formulas to remember

Sine: $sin A = \frac{opposite}{hypotenuse} = \frac{B C}{A C}$; Triangle right-angled at B; opposite is the side facing angle A.
Cosine: $cos A = \frac{adjacent}{hypotenuse} = \frac{A B}{A C}$; Adjacent is the side next to A that is not the hypotenuse.
Tangent: $tan A = \frac{opposite}{adjacent} = \frac{B C}{A B}$; Does not use the hypotenuse; can be more than 1.
Reciprocal ratios: $cosec A = \frac{A C}{B C}, sec A = \frac{A C}{A B}, cot A = \frac{A B}{B C}$; Flip sin, cos and tan respectively.
Pythagoras theorem: $A C^{2} = A B^{2} + B C^{2}$; Find the missing side before writing any ratio.
Well-definedness (similar right triangles): $∠ A = ∠ A^{'} \Rightarrow \frac{B C}{A C} = \frac{B ^{'} C ^{'}}{A ^{'} C ^{'}}$; AA similarity; so a ratio depends only on the angle. Use sides ak, bk.
Bounds for an acute angle: $0 < sin A < 1, 0 < cos A < 1, sec A > 1, cosec A > 1$; Use to reject impossible values such as sin A = 7/5.

How to solve Trigonometric Ratios of an Acute Angle questions

This method works for any question that gives a right triangle (or a figure containing one) and asks for ratios or sides.

1Identify the right angle and write the hypotenuse (the side opposite the right angle).
2Mark the angle asked about, then name its opposite and adjacent sides.
3If a side is missing, find it by Pythagoras; if a sum or difference of sides is given, factorise $h^{2} - s^{2} = (h - s) (h + s)$ .
4If only a ratio is known, write sides as multiples of $k$ ; the $k$ cancels later.
5Write each required ratio straight from the definition and simplify.
6If the figure has two right triangles sharing an angle, use that the ratio is equal in both to find unknown lengths.
7Check: sine and cosine of an acute angle must lie between 0 and 1, and include units for any length.

Trigonometric Ratios of an Acute Angle 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.

Trigonometric ratios of an acute angle

Imagine you are flying a kite on a school ground during a festival. The string is pulled straight and is 50 m long, and the kite is directly above a point on the ground 30 m from your hand (treat your hand as being at ground level).

Let $θ$ be the angle the string makes with the ground. Find the height of the kite, $sin θ$ and $tan θ$ . If you let out more string until it is 75 m long while keeping the same angle $θ$ , how high is the kite now and how far away horizontally is the point below it?

Try it first, then show the solutionSolution

Hypotenuse = string = 50 m, adjacent to $θ$ = 30 m.
Height $= 5 0^{2} - 3 0^{2} = 1600 = 40$ m.
$sin θ = \frac{40}{50} = \frac{4}{5}$ , $tan θ = \frac{40}{30} = \frac{4}{3}$ .
Same angle, so the ratios stay the same (ratios depend only on the angle).
New height $= 75 \times sin θ = 75 \times \frac{4}{5} = 60$ m.
New horizontal distance $= 75 \times cos θ = 75 \times \frac{3}{5} = 45$ m. Check: $4 5^{2} + 6 0^{2} = 5625 = 7 5^{2}$ .

Answer: Height 40 m, $sin θ = \frac{4}{5}$ , $tan θ = \frac{4}{3}$ ; with 75 m of string the kite is 60 m high and 45 m away horizontally.

Memory hook: Same kite angle, longer string: the triangle grows but sin and tan stay put.

Tricks to remember Trigonometric Ratios of an Acute Angle

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: Write it at the top of your rough work before labelling sides.
One 'co' per pair
sin pairs with cosec, cos with sec, tan with cot. Each reciprocal pair has exactly one 'co'.
Use it when: Stops you writing sec as 1/sin.
Stand at the angle
Imagine standing at the vertex: the side you face is opposite, the side under your feet is adjacent, the longest side is always the hypotenuse.
Use it when: Relabel every time the question switches from angle A to angle C.

Trigonometric Ratios of an Acute Angle: solved examples

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

Example 1: Ratios of both acute angles · 2 marks · Very short answerBoard level

In $△ P QR$ , right-angled at $R$ , $P Q = 25$ cm and $QR = 7$ cm. Find $sin P$ , $tan P$ and $cos Q$ .

Try it first, then show the solutionSolution

Hypotenuse is $P Q$ (opposite the right angle at $R$ ).
$P R = 2 5^{2} - 7^{2} = 576 = 24$ cm. (½ mark)
For $∠ P$ : opposite $= QR = 7$ , adjacent $= P R = 24$ .
$sin P = \frac{7}{25}$ , $tan P = \frac{7}{24}$ . (1 mark)
For $∠ Q$ : adjacent $= QR = 7$ , so $cos Q = \frac{7}{25}$ . (½ mark)

Answer: $sin P = \frac{7}{25}$ , $tan P = \frac{7}{24}$ , $cos Q = \frac{7}{25}$

Example 2: Sides from a difference condition · 3 marks · Short answerHarder board level

In $△ A B C$ , right-angled at $B$ , $B C = 6$ cm and $A C - A B = 2$ cm. Find $sin A$ , $cos A$ , $tan C$ and the value of $sec A + tan A$ .

Try it first, then show the solutionSolution

Pythagoras: $A C^{2} - A B^{2} = B C^{2} = 36$ .
$(A C - A B) (A C + A B) = 36$ , so $2 (A C + A B) = 36$ , giving $A C + A B = 18$ . (1 mark)
Adding with $A C - A B = 2$ : $A C = 10$ cm, $A B = 8$ cm. (½ mark)
$sin A = \frac{B C}{A C} = \frac{3}{5}$ , $cos A = \frac{A B}{A C} = \frac{4}{5}$ . (½ mark)
For $∠ C$ : opposite $= A B$ , adjacent $= B C$ , so $tan C = \frac{8}{6} = \frac{4}{3}$ . (½ mark)
$sec A + tan A = \frac{10}{8} + \frac{6}{8} = 2$ . (½ mark)

Answer: $sin A = \frac{3}{5}$ , $cos A = \frac{4}{5}$ , $tan C = \frac{4}{3}$ , $sec A + tan A = 2$

Example 3: Altitude to the hypotenuse · 3 marks · Short answerHarder board level

In $△ A B C$ , right-angled at $B$ , $A B = 15$ cm and $B C = 20$ cm. $B D$ is drawn perpendicular to $A C$ , meeting it at $D$ . Using trigonometric ratios of $∠ A$ in two triangles, find $B D$ and $A D$ .

Try it first, then show the solutionSolution

$A C = 1 5^{2} + 2 0^{2} = 25$ cm. (½ mark)
In $△ A B C$ : $sin A = \frac{20}{25} = \frac{4}{5}$ , $cos A = \frac{15}{25} = \frac{3}{5}$ . (1 mark)
$△ A B D$ is right-angled at $D$ and contains the same $∠ A$ , so the ratios are equal (ratios depend only on the angle).
$sin A = \frac{B D}{A B}$ , so $B D = 15 \times \frac{4}{5} = 12$ cm. (½ mark)
$cos A = \frac{A D}{A B}$ , so $A D = 15 \times \frac{3}{5} = 9$ cm. (½ mark)
Check: $9^{2} + 1 2^{2} = 225 = 1 5^{2}$ . (½ mark)

Answer: $B D = 12$ cm, $A D = 9$ cm

Example 4: Proof of well-definedness and its use · 5 marks · Long answerHarder board level

(i) $△ A B C$ and $△ D E F$ are right-angled at $B$ and $E$ respectively, with $∠ A = ∠ D$ . Prove that $sin A = sin D$ and $tan A = tan D$ .
(ii) In $△ A B C$ , $A B = 8$ cm and $B C = 6$ cm. In $△ D E F$ , the hypotenuse $D F = 25$ cm. Find $E F$ , $D E$ and $cot D$ .

Try it first, then show the solutionSolution

(i) $∠ B = ∠ E = 9 0^{\circ}$ and $∠ A = ∠ D$ (given).
So $△ A B C \sim △ D E F$ (AA similarity). (1 mark)
Corresponding sides are proportional: $\frac{B C}{E F} = \frac{A C}{D F} = \frac{A B}{D E}$ .
Rearranging, $\frac{B C}{A C} = \frac{E F}{D F}$ , i.e. $sin A = sin D$ ; and $\frac{B C}{A B} = \frac{E F}{D E}$ , i.e. $tan A = tan D$ . (1 mark)
(ii) $A C = 8^{2} + 6^{2} = 10$ cm, so $sin A = \frac{3}{5}$ , $cos A = \frac{4}{5}$ . (1 mark)
$E F = D F sin D = 25 \times \frac{3}{5} = 15$ cm. (½ mark)
$D E = D F cos D = 25 \times \frac{4}{5} = 20$ cm. (½ mark)
$cot D = \frac{D E}{E F} = \frac{20}{15} = \frac{4}{3}$ ; check $cot A = \frac{8}{6} = \frac{4}{3}$ . (1 mark)

Answer: (i) Proved by AA similarity. (ii) $E F = 15$ cm, $D E = 20$ cm, $cot D = \frac{4}{3}$

Example 5: Case study: a school ramp · 4 marks · Case-basedHOTS / competency

Suppose your school builds a straight ramp up to a stage that is 0.6 m high. The first design has its foot 0.8 m from the stage, measured along level ground. Let $θ$ be the angle the ramp makes with the ground. Suppose the school's own safety rule says $tan θ$ must not be more than $\frac{5}{12}$ .
(i) Find the length of the first ramp. (1 mark)
(ii) Find $tan θ$ for the first ramp and say whether it meets the rule. (1 mark)
(iii) For a ramp to the same 0.6 m stage with $tan θ = \frac{5}{12}$ exactly, find the horizontal distance, the ramp length and $sin θ$ . (2 marks)

Try it first, then show the solutionSolution

(i) Length $= 0. 6^{2} + 0. 8^{2} = 1 = 1$ m. (1 mark)
(ii) $tan θ = \frac{0.6}{0.8} = \frac{3}{4}$ ; since $\frac{3}{4} = \frac{9}{12} > \frac{5}{12}$ , it does not meet the rule. (1 mark)
(iii) $\frac{0.6}{x} = \frac{5}{12}$ , so $x = \frac{0.6 \times 12}{5} = 1.44$ m. (½ mark)
Sides are in ratio $5 : 12$ , so hypotenuse ratio is 13: length $= 0.12 \times 13 = 1.56$ m. (1 mark)
$sin θ = \frac{0.6}{1.56} = \frac{5}{13}$ . (½ mark)

Answer: (i) 1 m (ii) $tan θ = \frac{3}{4}$ , too steep (iii) 1.44 m, 1.56 m, $sin θ = \frac{5}{13}$

Where students lose marks in Trigonometric Ratios of an Acute Angle

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

Mistake: Using the wrong side as opposite
The labels are taken from the wrong angle, often the other acute angle.
Wrong working
Right angle at $B$ , $A B = 3$ , $B C = 4$ : $sin A = \frac{3}{5}$
Right working
Opposite $A$ is $B C$ : $sin A = \frac{4}{5}$
Do this: Mark the angle, then write 'opp = ..., adj = ...' before any ratio.
Usually costs: The whole answer and every later step.
Mistake: Adding squares when the hypotenuse is given
Pythagoras applied by habit as a sum.
Wrong working
Hypotenuse 13, one side 5: other side $= 169 + 25$
Right working
Other side $= 169 - 25 = 12$
Do this: Hypotenuse squared is the larger term; subtract to find a leg.
Usually costs: ½ to 1 mark and wrong ratios after.
Mistake: Treating a ratio as actual side lengths
Forgetting that ratios fix only proportions.
Wrong working
$sin A = \frac{3}{5}$ , so $B C = 3$ cm
Right working
$B C = 3 k$ , $A C = 5 k$ for some $k > 0$
Do this: Use $k$ unless a length is actually given.
Usually costs: Marks for justification in 2- and 3-mark answers.
Mistake: Cancelling the angle letter
Reading $sin A$ as $sin \times A$ .
Wrong working
$\frac{s i n A}{s i n B} = \frac{A}{B}$
Right working
$sin A$ is one number; it cannot be split.
Do this: Treat each ratio as a single symbol.
Usually costs: Full marks for the step.

Exam tips

Write the hypotenuse and opposite/adjacent sides for the named angle in your working; examiners give step marks for it.
When a sum or difference of two sides is given, factorise the difference of squares; it gives a linear equation quickly.
In figures with an altitude, remember the same angle sits in two right triangles, so its ratio is equal in both.
Simplify ratios fully and give units for lengths, not for ratios.
For assertion-reason items, test the claim with an actual triangle such as 3, 4, 5.

Practice questions on Trigonometric Ratios of an Acute Angle

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
In $△ A B C$ , right-angled at $C$ , $A B = 17$ cm and $B C = 8$ cm. The value of $tan A$ is
(a) $\frac{8}{17}$
(b) $\frac{15}{8}$
(c) $\frac{8}{15}$
(d) $\frac{15}{17}$
Hint
Find $A C$ first; for $∠ A$ the opposite side is $B C$ .
Answer
(c) $\frac{8}{15}$ . $A C = 289 - 64 = 15$ , $tan A = \frac{B C}{A C} = \frac{8}{15}$ .
Q2 · 3 marks · Short answerHarder board level
In $△ X Y Z$ , right-angled at $Y$ , $X Y = 7$ cm and $X Z - Y Z = 1$ cm. Find $sin X$ , $cos Z$ and the value of $sec X + tan X$ .
Hint
$X Z^{2} - Y Z^{2} = 49$ ; factorise.
Answer
$(X Z - Y Z) (X Z + Y Z) = 49$ gives $X Z + Y Z = 49$ , so $X Z = 25$ , $Y Z = 24$ . $sin X = \frac{24}{25}$ , $cos Z = \frac{Y Z}{X Z} = \frac{24}{25}$ , $sec X + tan X = \frac{25}{7} + \frac{24}{7} = 7$ .
Q3 · 3 marks · Short answerHarder board level
In $△ A B C$ , right-angled at $B$ , $A B = 12$ cm and $B C = 16$ cm. $D$ is the mid-point of $B C$ . Find $sin ∠ A D B$ , $tan ∠ B A D$ and $tan ∠ B A C$ .
Hint
Work in right $△ A B D$ with $B D = 8$ cm.
Answer
$A D = 144 + 64 = 413$ cm. $sin ∠ A D B = \frac{12}{4 13} = \frac{3}{13}$ , $tan ∠ B A D = \frac{8}{12} = \frac{2}{3}$ , $tan ∠ B A C = \frac{16}{12} = \frac{4}{3}$ .
Q4 · 5 marks · Long answerHarder board level
In $△ P QR$ , right-angled at $Q$ , $QS$ is perpendicular to $P R$ , with $P S = 9$ cm and $S R = 16$ cm. Find $QS$ , $P Q$ and $QR$ . Then find $sin P$ in both $△ P QS$ and $△ P QR$ , and find $tan R$ .
Hint
$△ P S Q \sim △ QS R$ gives $Q S^{2} = P S \cdot S R$ .
Answer
$Q S^{2} = 9 \times 16 = 144$ , $QS = 12$ cm. $P Q = 81 + 144 = 15$ cm, $QR = 144 + 256 = 20$ cm, $P R = 25$ cm. In $△ P QS$ : $sin P = \frac{12}{15} = \frac{4}{5}$ ; in $△ P QR$ : $sin P = \frac{20}{25} = \frac{4}{5}$ (same, as expected). $tan R = \frac{P Q}{QR} = \frac{3}{4}$ .
Q5 · 3 marks · Short answerHOTS / competency
Riya solves: "In $△ A B C$ , right-angled at $B$ , $A B = 5$ cm and $B C = 12$ cm, so $sin A = \frac{5}{13}$ . If every side is tripled, $sin A$ becomes $\frac{15}{39}$ , which is a different value." Find both errors and give the correct $sin A$ and $cos A$ .
Hint
Which side faces $∠ A$ ? Is $\frac{15}{39}$ really different?
Answer
Error 1: she used the adjacent side $A B$ . $A C = 13$ , so $sin A = \frac{B C}{A C} = \frac{12}{13}$ and $cos A = \frac{5}{13}$ . Error 2: tripling all sides keeps the angles, so the ratio is unchanged; any ratio such as $\frac{36}{39} = \frac{12}{13}$ simplifies to the same value.

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 $tan A$ in a right triangle.Show answer
Opposite side over adjacent side of angle $A$ .
2.Reciprocal of $cos A$ ?Show answer
$sec A$
3.Why can $sin A$ of an acute angle never exceed 1?Show answer
The hypotenuse is the longest side, so opposite/hypotenuse is less than 1.
4.Which similarity criterion proves the ratios are well defined?Show answer
AA similarity.
5.Right angle at $B$ : which side is the hypotenuse?Show answer
$A C$

Trigonometric Ratios of an Acute Angle: frequently asked questions

Do trigonometric ratios change if the triangle is bigger?

No. Two right triangles with the same acute angle are similar, so their sides are in the same proportion. The ratios depend only on the angle.

Can tan A be greater than 1?

Yes. Tangent compares the opposite side with the adjacent side, and the opposite side can be longer. Only sine and cosine of an acute angle stay below 1.

Why do we write sides as 3k and 5k?

A ratio only fixes the proportion of sides, not their actual lengths. The constant k shows any size is possible, and it cancels when you form another ratio.

Are trigonometric ratios of complementary angles in the syllabus?

No, trigonometric ratios of complementary angles as a topic are not in the 2026-27 syllabus.