MathsCBSE Class 10 · Introduction to Trigonometry

Finding Trigonometric Ratios from a Given Ratio

Q: Do I always need to use k when finding ratios?

Writing k shows the examiner you know a ratio is not a length, and it is essential when a real side is given. If only ratios are asked, k cancels, but writing it costs nothing and protects your marks.

Q: Which ratio should I find first if the question gives tan?

Tan gives you the opposite and adjacent sides, so find the hypotenuse by Pythagoras first. Then sin and cos follow directly, and the other three are their reciprocals.

Q: Why are all my ratios positive?

In this chapter the angle is an acute angle of a right triangle, and every side length is positive. So every ratio is positive and you take only the positive square root.

Q: Is the topic of complementary angles, like sin of 90 minus A, needed for this?

No. Trigonometric ratios of complementary angles are not in the 2026-27 syllabus. You only need the definitions, Pythagoras and the basic relations for this topic.

Q: How can I check my answer quickly in the exam?

Square your sine and cosine values and add them; the sum must be 1. Also make sure sin and cos are less than 1 and sec and cosec are greater than 1.

Updated 13 October 2026 · Fact-checked

Finding ratios from a given ratio means using one known trigonometric ratio of an acute angle to get all the others. You write the given ratio as two sides with a constant k, find the third side by Pythagoras, then read off the rest. The board asks for missing ratios, values of expressions, and real side lengths.

Understand Finding Ratios from a Given Ratio

Suppose you are told only that $sin A = \frac{3}{5}$ for an acute angle $A$ . That one number is enough to find $cos A$ , $tan A$ and every other ratio. This topic teaches you how.

The idea comes from the definition. $sin A = \frac{3}{5}$ means the side opposite $A$ and the hypotenuse are in the ratio $3 : 5$ . It does not mean they are exactly 3 and 5 units long. Any right triangle with that angle $A$ has the same ratio, because all such triangles are similar. So you write opposite $= 3 k$ and hypotenuse $= 5 k$ , where $k$ is some positive number.

Now Pythagoras gives the missing side: adjacent $= 25 k^{2} - 9 k^{2} = 4 k$ . Every other ratio is one side divided by another, so the $k$ cancels: $cos A = \frac{4 k}{5 k} = \frac{4}{5}$ and $tan A = \frac{3}{4}$ . You always get the same answer, whatever the size of the triangle.

The $k$ matters when the question also gives a real length. If the hypotenuse is actually 20 cm and $sin A = \frac{3}{5}$ , then $5 k = 20$ , so $k = 4$ and the opposite side is 12 cm. Board questions often hide a real length like this, or give a sum or difference of two sides that you must turn into $k$ first.

This topic uses three things from the rest of the chapter: the side labels (opposite, adjacent, hypotenuse), the reciprocal and quotient relations, and Pythagoras. You can also check your work with $sin^{2} A + cos^{2} A = 1$ . For $\frac{3}{5}$ and $\frac{4}{5}$ : $\frac{9}{25} + \frac{16}{25} = 1$ .

The method works only for an acute angle in a right triangle. Then all six ratios are positive, so you take only the positive square root. Also, $sin A$ and $cos A$ must lie between 0 and 1, and $sec A$ and $cosec A$ must be greater than 1. A given value like $cos A = \frac{7}{5}$ is impossible, because the hypotenuse is the longest side.

Finding Ratios from a Given Ratio: formulas to remember

Writing a given ratio with a constant k: $sin A = \frac{p}{h} \Rightarrow opposite = p k, hypotenuse = hk, k > 0$; Do the same for any ratio: tan A = p/b gives opposite pk and adjacent bk.
Third side by Pythagoras: $hypotenuse^{2} = opposite^{2} + adjacent^{2}$; Take only the positive root; lengths are positive.
Difference of squares form: $h^{2} - s^{2} = (h - s) (h + s)$; Use when the sum or difference of the hypotenuse and a side is given.
Reciprocal relations: $cosec A = \frac{1}{s i n A}, sec A = \frac{1}{c o s A}, cot A = \frac{1}{t a n A}$; Once sin, cos and tan are found, flip them for the other three.
Divide-by-cos shortcut: $\frac{a s i n θ + b c o s θ}{c s i n θ + d c o s θ} = \frac{a t a n θ + b}{c t a n θ + d}$; Works only when every term has exactly one sin or cos (no stand-alone numbers).
Check identity: $sin^{2} A + cos^{2} A = 1$; Substitute your two values to confirm the working.

How to solve Finding Ratios from a Given Ratio questions

This method handles every board question that gives one ratio and asks for others, an expression, or a real side length.

1Draw a right triangle, mark the right angle and the angle in the question, and label opposite, adjacent and hypotenuse from that angle.
2If the ratio is given in a disguised form (like $4 tan A = 3$ or $7 sin A = 24 cos A$ ), first rewrite it as a single ratio, such as $tan A = \frac{3}{4}$ .
3Write the two sides it names as $p k$ and $q k$ with $k > 0$ .
4Find the third side by Pythagoras, taking the positive root. If the sum or difference of two sides is given instead, use $(h - s) (h + s)$ .
5If a real length is given, equate it to the matching $k$ -side, find $k$ , and get actual lengths with units.
6Write the required ratios as side over side; the $k$ cancels.
7For an expression, substitute the values, simplify numerator and denominator separately, then divide.
8Check with $sin^{2} A + cos^{2} A = 1$ or the divide-by-cos shortcut.

Finding Ratios from a Given Ratio 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.

Finding all ratios from one given ratio

Imagine your school ground is a rectangle 120 m long. Students always cut straight across it along the diagonal to reach the canteen. Suppose the angle $θ$ between this diagonal path and the longer edge has $cos θ = \frac{15}{17}$ .

Find the width of the ground and the length of the diagonal path. How many metres does a student save by walking along the diagonal instead of along the length and then the width?

Try it first, then show the solutionSolution

$cos θ = \frac{adjacent}{hypotenuse}$ , so length $= 15 k$ and diagonal $= 17 k$ .
Width $= 289 k^{2} - 225 k^{2} = 64 k^{2} = 8 k$ .
Length is 120 m, so $15 k = 120$ and $k = 8$ .
Width $= 8 \times 8 = 64$ m and diagonal $= 17 \times 8 = 136$ m.
Walking along the edges: $120 + 64 = 184$ m.
Saving $= 184 - 136 = 48$ m.
Check: $12 0^{2} + 6 4^{2} = 14400 + 4096 = 18496 = 13 6^{2}$ .

Answer: Width 64 m, diagonal 136 m; the student saves 48 m.

Memory hook: The shortcut across the ground: one ratio plus one real length gives you k, and k gives you every side.

Tricks to remember Finding Ratios from a Given Ratio

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

Never forget the k
Say it as "ratio, not length": $\frac{3}{5}$ becomes $3 k$ and $5 k$ , never 3 and 5.
Use it when: Saves you whenever a question gives a real side length along with a ratio.
Pythagorean triples to spot fast
Remember 3-4-5, 5-12-13, 8-15-17, 7-24-25, 20-21-29. If two numbers from a triple appear, the third side is the missing one.
Use it when: Lets you write the third side quickly and check your square root.
Hypotenuse is the biggest
In sin and cos the hypotenuse sits at the bottom, so the fraction is less than 1; in sec and cosec it sits on top, so the value is more than 1.
Use it when: Instantly rejects impossible values and catches a flipped ratio.

Finding Ratios from a Given Ratio: solved examples

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

Example 1: All ratios from a disguised cot · 2 marks · Very short answerBoard level

If $15 cot A = 8$ , where $A$ is acute, find $sin A$ , $cos A$ and $sec A$ .

Try it first, then show the solutionSolution

$cot A = \frac{8}{15} = \frac{adjacent}{opposite}$ , so adjacent $= 8 k$ , opposite $= 15 k$ . (½ mark)
Hypotenuse $= 64 k^{2} + 225 k^{2} = 289 k^{2} = 17 k$ . (½ mark)
$sin A = \frac{15 k}{17 k} = \frac{15}{17}$ and $cos A = \frac{8}{17}$ . (½ mark)
$sec A = \frac{1}{c o s A} = \frac{17}{8}$ . (½ mark)
Check: $\frac{225}{289} + \frac{64}{289} = 1$ .

Answer: $sin A = \frac{15}{17}$ , $cos A = \frac{8}{17}$ , $sec A = \frac{17}{8}$

Example 2: Expression that needs actual values · 3 marks · Short answerHarder board level

If $3 tan θ = 4$ , where $θ$ is acute, evaluate $\frac{4 s i n θ - c o s θ + 1}{4 s i n θ + c o s θ - 1}$ .

Try it first, then show the solutionSolution

$tan θ = \frac{4}{3}$ , so opposite $= 4 k$ , adjacent $= 3 k$ , hypotenuse $= 16 k^{2} + 9 k^{2} = 5 k$ . (1 mark)
$sin θ = \frac{4}{5}$ and $cos θ = \frac{3}{5}$ . (½ mark)
Numerator $= \frac{16}{5} - \frac{3}{5} + 1 = \frac{13}{5} + \frac{5}{5} = \frac{18}{5}$ . (½ mark)
Denominator $= \frac{16}{5} + \frac{3}{5} - 1 = \frac{19}{5} - \frac{5}{5} = \frac{14}{5}$ . (½ mark)
Value $= \frac{18}{5} \div \frac{14}{5} = \frac{18}{14} = \frac{9}{7}$ . (½ mark)
Note: the divide-by-cos shortcut does not apply here because of the stand-alone 1.

Answer: $\frac{9}{7}$

Example 3: Sides from a difference, then ratios · 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 $sin X$ , $cos X$ and $tan Z$ .

Try it first, then show the solutionSolution

By Pythagoras, $X Z^{2} - Y Z^{2} = X Y^{2} = 49$ . (½ mark)
$(X Z - Y Z) (X Z + Y Z) = 49$ , so $1 \times (X Z + Y Z) = 49$ , giving $X Z + Y Z = 49$ . (½ mark)
Adding $X Z - Y Z = 1$ and $X Z + Y Z = 49$ : $2 X Z = 50$ , so $X Z = 25$ cm and $Y Z = 24$ cm. (1 mark)
For $∠ X$ : opposite $= Y Z = 24$ , adjacent $= X Y = 7$ . So $sin X = \frac{24}{25}$ , $cos X = \frac{7}{25}$ . (½ mark)
For $∠ Z$ : opposite $= X Y = 7$ , adjacent $= Y Z = 24$ . So $tan Z = \frac{7}{24}$ . (½ mark)
Check: $7^{2} + 2 4^{2} = 49 + 576 = 625 = 2 5^{2}$ .

Answer: $X Z = 25$ cm, $Y Z = 24$ cm; $sin X = \frac{24}{25}$ , $cos X = \frac{7}{25}$ , $tan Z = \frac{7}{24}$

Example 4: Ratios in letters and a general result · 5 marks · Long answerHarder board level

If $sin θ = \frac{a}{a ^{2} + b ^{2}}$ , where $a > 0$ , $b > 0$ and $θ$ is acute:
(i) find $cos θ$ , $tan θ$ and $sec θ$ ;
(ii) prove that $\frac{a s i n θ - b c o s θ}{a s i n θ + b c o s θ} = \frac{a ^{2} - b ^{2}}{a ^{2} + b ^{2}}$ ;
(iii) verify (ii) when $a = 3$ and $b = 4$ by finding $sin θ$ and $cos θ$ directly.

Try it first, then show the solutionSolution

Opposite $= ak$ , hypotenuse $= a^{2} + b^{2} k$ . (½ mark)
Adjacent $= (a^{2} + b^{2}) k^{2} - a^{2} k^{2} = b^{2} k^{2} = bk$ . (½ mark)
$cos θ = \frac{b}{a ^{2} + b ^{2}}$ , $tan θ = \frac{a}{b}$ , $sec θ = \frac{a ^{2} + b ^{2}}{b}$ . (1 mark)
(ii) Divide numerator and denominator of the LHS by $cos θ$ : LHS $= \frac{a t a n θ - b}{a t a n θ + b}$ . (½ mark)
Substitute $tan θ = \frac{a}{b}$ : LHS $= \frac{\frac{a ^{2}}{b} - b}{\frac{a ^{2}}{b} + b}$ . (½ mark)
Multiply top and bottom by $b$ : LHS $= \frac{a ^{2} - b ^{2}}{a ^{2} + b ^{2}} =$ RHS. (1 mark)
(iii) With $a = 3$ , $b = 4$ : $sin θ = \frac{3}{5}$ , $cos θ = \frac{4}{5}$ . (½ mark)
LHS $= \frac{\frac{9}{5} - \frac{16}{5}}{\frac{9}{5} + \frac{16}{5}} = \frac{- 7}{25}$ and RHS $= \frac{9 - 16}{9 + 16} = \frac{- 7}{25}$ . Verified. (½ mark)

Answer: $cos θ = \frac{b}{a ^{2} + b ^{2}}$ , $tan θ = \frac{a}{b}$ , $sec θ = \frac{a ^{2} + b ^{2}}{b}$ ; the result is proved, and both sides equal $- \frac{7}{25}$ for $a = 3$ , $b = 4$ .

Example 5: Case study: flying a kite · 4 marks · Case-basedHOTS / competency

Imagine you are flying a kite in an open field. You hold the string at ground level and it is pulled straight and taut, 75 m long. The angle $θ$ the string makes with the level ground satisfies $sin θ = \frac{24}{25}$ .
(i) Find $cos θ$ . (1 mark)
(ii) Find the height of the kite above the ground. (1 mark)
(iii) You let out more string until it is 100 m long, and the string keeps the same angle with the ground. Find the new horizontal distance of the kite from you and how much higher the kite has risen. (2 marks)

Try it first, then show the solutionSolution

Height $= 24 k$ , string $= 25 k$ , horizontal distance $= 625 k^{2} - 576 k^{2} = 7 k$ .
(i) $cos θ = \frac{7}{25}$ . (1 mark)
(ii) $25 k = 75$ , so $k = 3$ and height $= 24 \times 3 = 72$ m. (1 mark)
(iii) Same angle, so the ratio stays $7 : 24 : 25$ ; now $25 k = 100$ , so $k = 4$ . (½ mark)
New horizontal distance $= 7 \times 4 = 28$ m. (½ mark)
New height $= 24 \times 4 = 96$ m, so rise $= 96 - 72 = 24$ m. (1 mark)

Answer: (i) $cos θ = \frac{7}{25}$ (ii) 72 m (iii) horizontal distance 28 m; the kite rises 24 m higher.

Where students lose marks in Finding Ratios from a Given Ratio

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

Mistake: Taking the ratio as the actual side lengths
The ratio gives only the proportion of the sides. When a real length is given, the answer comes out wrong.
Wrong working
$tan A = \frac{3}{4}$ and hypotenuse $= 15$ cm, so opposite $= 3$ cm.
Right working
Opposite $= 3 k$ , adjacent $= 4 k$ , hypotenuse $= 5 k = 15$ , so $k = 3$ and opposite $= 9$ cm.
Do this: Always write sides as $p k$ and $q k$ , then find $k$ from the real length.
Usually costs: The length marks and every result that uses that length.
Mistake: Labelling opposite and adjacent from the wrong angle
When a question asks about $∠ C$ after you worked with $∠ A$ , the opposite and adjacent sides swap.
Wrong working
In $△ A B C$ right-angled at $B$ with $A B = 5$ , $B C = 12$ : $tan C = \frac{12}{5}$ .
Right working
Opposite $C$ is $A B = 5$ and adjacent is $B C = 12$ , so $tan C = \frac{5}{12}$ .
Do this: Before each ratio, put your finger on the angle and name the side across from it.
Usually costs: ½ to 1 mark per wrong ratio, and the expression built from it.
Mistake: Finding the third side by adding or subtracting instead of squaring
Pythagoras works with squares; sides do not add like this.
Wrong working
$sin A = \frac{5}{13}$ , so adjacent $= 13 - 5 = 8$ .
Right working
Adjacent $= 169 k^{2} - 25 k^{2} = 144 k^{2} = 12 k$ .
Do this: Write the Pythagoras line every time, then check with a known triple.
Usually costs: Nearly all marks, since every later ratio is wrong.
Mistake: Using the divide-by-cos shortcut when a plain number is present
Dividing by $cos θ$ turns a stand-alone 1 into $sec θ$ , not 1.
Wrong working
$\frac{4 s i n θ - c o s θ + 1}{4 s i n θ + c o s θ - 1} = \frac{4 t a n θ - 1 + 1}{4 t a n θ + 1 - 1}$
Right working
Substitute the actual values $sin θ = \frac{4}{5}$ and $cos θ = \frac{3}{5}$ and simplify, giving $\frac{9}{7}$ .
Do this: Use the shortcut only when every term has exactly one sin or cos.
Usually costs: The evaluation marks, usually 1 to 2 marks.

Exam tips

Show the Pythagoras line with $k$ clearly; examiners give step marks for the third side even if a later step slips.
Read whether the angle is acute. Then write all ratios as positive and take only the positive root.
When the question gives a sum or difference of sides, turn it into $(h - s) (h + s)$ first; this is the usual 3-mark version.
In MCQs, check your answer quickly with $sin^{2} A + cos^{2} A = 1$ or by rejecting values greater than 1 for sin and cos.
In case studies, state units for every length and find $k$ from the real length before answering each sub-part.

Practice questions on Finding Ratios from a Given Ratio

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

Q1 · 2 marks · Very short answerBoard level
If $7 sin A = 24 cos A$ , where $A$ is acute, find the value of $sin A + cos A$ .
Hint
Divide both sides by $7 cos A$ to get $tan A$ first.
Answer
$tan A = \frac{24}{7}$ , so opposite $24 k$ , adjacent $7 k$ , hypotenuse $25 k$ . $sin A + cos A = \frac{24}{25} + \frac{7}{25} = \frac{31}{25}$ .
Q2 · 3 marks · Short answerHarder board level
If $cos θ = 0.6$ , where $θ$ is acute, evaluate $\frac{5 s i n θ - 3 t a n θ}{2 s e c θ + 4 c o t θ}$ .
Hint
Write 0.6 as a fraction in lowest terms, then use a 3-4-5 triangle.
Answer
$cos θ = \frac{3}{5}$ , so $sin θ = \frac{4}{5}$ , $tan θ = \frac{4}{3}$ , $sec θ = \frac{5}{3}$ , $cot θ = \frac{3}{4}$ . Numerator $= 4 - 4 = 0$ ; denominator $= \frac{10}{3} + 3 = \frac{19}{3}$ . Value $= 0$ .
Q3 · 3 marks · Short answerHarder board level
In $△ A B C$ , right-angled at $C$ , $B C = 12$ cm and $A B - A C = 8$ cm. Find $sin A$ , $cos A$ and the value of $sin^{2} A - cos^{2} A$ .
Hint
$A B$ is the hypotenuse. Use $(A B - A C) (A B + A C) = B C^{2}$ .
Answer
$(A B - A C) (A B + A C) = 144$ , so $A B + A C = 18$ . Then $A B = 13$ cm, $A C = 5$ cm. $sin A = \frac{12}{13}$ , $cos A = \frac{5}{13}$ , and $sin^{2} A - cos^{2} A = \frac{144 - 25}{169} = \frac{119}{169}$ .
Q4 · 5 marks · Long answerHarder board level
If $sec θ = \frac{5}{4}$ , where $θ$ is acute:
(i) find all the other trigonometric ratios of $θ$ ;
(ii) show that $\frac{t a n θ}{1 + t a n ^{2} θ} = \frac{s i n θ}{s e c θ}$ ;
(iii) find the value of $\frac{1 - s i n θ}{1 + s i n θ}$ and of $cot θ + cosec θ$ .
Hint
$sec θ = \frac{hypotenuse}{adjacent}$ , so take hypotenuse $5 k$ and adjacent $4 k$ .
Answer
(i) Opposite $= 3 k$ . $sin θ = \frac{3}{5}$ , $cos θ = \frac{4}{5}$ , $tan θ = \frac{3}{4}$ , $cosec θ = \frac{5}{3}$ , $cot θ = \frac{4}{3}$ . (ii) LHS $= \frac{3}{4} \div \frac{25}{16} = \frac{12}{25}$ ; RHS $= \frac{3}{5} \times \frac{4}{5} = \frac{12}{25}$ , so equal. (iii) $\frac{2/5}{8/5} = \frac{1}{4}$ ; $cot θ + cosec θ = \frac{4}{3} + \frac{5}{3} = 3$ .
Q5 · 4 marks · Case-basedHOTS / competency
In $△ A B C$ , right-angled at $B$ , $sin A = \frac{2}{3}$ and $A C = 12$ cm. Ravi writes: " $sin A = \frac{2}{3}$ , so $B C = 2$ cm."
(i) Explain Ravi's error.
(ii) Find the correct lengths of $B C$ and $A B$ , and the perimeter of the triangle.
(iii) Could $cos A$ be $\frac{3}{2}$ in some right triangle? Give a reason.
Hint
A ratio fixes only the proportion of the sides. Bring in $k$ and use the real hypotenuse.
Answer
(i) $\frac{2}{3}$ is a ratio, so $B C = 2 k$ , $A C = 3 k$ ; it does not give actual lengths. (ii) $3 k = 12$ , $k = 4$ , $B C = 8$ cm, $A B = 144 - 64 = 80 = 45$ cm; perimeter $= 20 + 45$ cm. (iii) No: the adjacent side would be longer than the hypotenuse, but the hypotenuse is the longest side, so $cos A < 1$ for an acute angle.

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.If $sin A = \frac{3}{5}$ , what is $cos A$ ?Show answer
$\frac{4}{5}$
2.If $tan A = \frac{4}{3}$ , what is $sec A$ ?Show answer
$\frac{5}{3}$
3.If $cos A = \frac{12}{13}$ , what is $tan A$ ?Show answer
$\frac{5}{12}$
4.Why do we write sides as $3 k$ and $5 k$ and not 3 and 5?Show answer
A ratio fixes only the proportion of the sides; similar triangles of any size share it.
5.Can $sin A = \frac{5}{4}$ for an acute angle?Show answer
No. The hypotenuse is the longest side, so $sin A < 1$ .
6.If $cosec A = \frac{25}{7}$ , what is $cot A$ ?Show answer
$\frac{24}{7}$

Finding Ratios from a Given Ratio: frequently asked questions

Do I always need to use k when finding ratios?

Writing k shows the examiner you know a ratio is not a length, and it is essential when a real side is given. If only ratios are asked, k cancels, but writing it costs nothing and protects your marks.

Which ratio should I find first if the question gives tan?

Tan gives you the opposite and adjacent sides, so find the hypotenuse by Pythagoras first. Then sin and cos follow directly, and the other three are their reciprocals.

Why are all my ratios positive?

In this chapter the angle is an acute angle of a right triangle, and every side length is positive. So every ratio is positive and you take only the positive square root.

Is the topic of complementary angles, like sin of 90 minus A, needed for this?

No. Trigonometric ratios of complementary angles are not in the 2026-27 syllabus. You only need the definitions, Pythagoras and the basic relations for this topic.

How can I check my answer quickly in the exam?

Square your sine and cosine values and add them; the sum must be 1. Also make sure sin and cos are less than 1 and sec and cosec are greater than 1.