One Given Ratio Is Enough to Unlock All Six for the Same Angle
Write the six trigonometric ratios of an acute angle from the sides of a right triangle, recover the other five from any one of them using Pythagoras, learn the values for zero, thirty, forty-five, sixty and ninety degrees, and solve equations for an unknown angle.
Why does the ratio of two sides depend only on the angle?
Draw a right triangle with an acute angle of . Now draw a much larger one, also with a angle. The two triangles are similar, by the AA criterion, because both have a right angle and both have a angle — so the third angles match too.
And similar triangles have proportional sides. **So the ratio of the side opposite the angle to the hypotenuse is the same in both triangles**, and in every other right triangle containing a angle, however large or small.
That is the fact the whole of trigonometry rests on. A ratio of two sides of a right triangle is a property of the angle alone, not of the particular triangle — which is why it makes sense to give that ratio a name and a value.
There are six such ratios, because there are three sides and you can pick two of them in six ordered ways. Three have names you will use constantly and three are their reciprocals.
From there this part of the chapter does four things.
- Compute all six ratios from the three side lengths of a right triangle
- Recover the other five when only one is given, using the Pythagoras theorem to find the missing side
- Learn the exact values for , , , and , and evaluate expressions built from them
- Solve for an unknown angle from an equation such as
One thing to be clear about from the start. In , the "" is not a quantity being multiplied by — **it is the name of an operation applied to the angle .** Writing is meaningless, and "" on its own has no value at all.
This page covers the first part of the CBSE Class 10 Maths chapter on trigonometry: the six ratios, finding all of them from one, the standard angles, and solving for an angle.
And similar triangles have proportional sides. **So the ratio of the side opposite the angle to the hypotenuse is the same in both triangles**, and in every other right triangle containing a angle, however large or small.
That is the fact the whole of trigonometry rests on. A ratio of two sides of a right triangle is a property of the angle alone, not of the particular triangle — which is why it makes sense to give that ratio a name and a value.
There are six such ratios, because there are three sides and you can pick two of them in six ordered ways. Three have names you will use constantly and three are their reciprocals.
From there this part of the chapter does four things.
- Compute all six ratios from the three side lengths of a right triangle
- Recover the other five when only one is given, using the Pythagoras theorem to find the missing side
- Learn the exact values for , , , and , and evaluate expressions built from them
- Solve for an unknown angle from an equation such as
One thing to be clear about from the start. In , the "" is not a quantity being multiplied by — **it is the name of an operation applied to the angle .** Writing is meaningless, and "" on its own has no value at all.
This page covers the first part of the CBSE Class 10 Maths chapter on trigonometry: the six ratios, finding all of them from one, the standard angles, and solving for an angle.
Formula
What are the six trigonometric ratios of an acute angle?
Three ratios and their three reciprocals, all measured from the angle you are working with.
In a right triangle, for an acute angle , label the sides relative to : the opposite side faces , the adjacent side touches and is not the hypotenuse, and the hypotenuse faces the right angle.
Two relations follow immediately from the definitions, and both are examined:
Worked example 1. In , right-angled at , cm and cm. Find all six ratios of angle .
First find the hypotenuse by the Pythagoras theorem:
**Now identify the sides relative to .** The side opposite is , the side adjacent to is , and the hypotenuse is . So
Two checks worth running. First, . Correct. Second,
**Exactly ** — which it must be, since is the Pythagoras theorem divided by . That single check catches almost every error in this section.
Worked example 2 — the ratios of the other acute angle. For the same triangle, find and .
Relative to , the opposite side is and the adjacent side is , with the same hypotenuse:
Notice what happened. and . The two acute angles of a right triangle swap the roles of opposite and adjacent, which is why the sine of one equals the cosine of the other — and since , that is the beginning of the complementary-angle relations you meet in Part 2.
The mistake this example exists to prevent. "Opposite" and "adjacent" are not fixed labels on the triangle — they depend entirely on which angle you are working with. **The same side is opposite and adjacent to . A student who labels the triangle once and uses those labels for both angles will get half the answers wrong, and the error is invisible in the working.
Why the hypotenuse is special.** It is the longest side, so both and are always **less than ** for an acute angle, while and are always **greater than . And can be anything positive**, since neither of its sides is the hypotenuse. Those bounds are worth remembering as a first check on any answer.
In a right triangle, for an acute angle , label the sides relative to : the opposite side faces , the adjacent side touches and is not the hypotenuse, and the hypotenuse faces the right angle.
Two relations follow immediately from the definitions, and both are examined:
Worked example 1. In , right-angled at , cm and cm. Find all six ratios of angle .
First find the hypotenuse by the Pythagoras theorem:
**Now identify the sides relative to .** The side opposite is , the side adjacent to is , and the hypotenuse is . So
Two checks worth running. First, . Correct. Second,
**Exactly ** — which it must be, since is the Pythagoras theorem divided by . That single check catches almost every error in this section.
Worked example 2 — the ratios of the other acute angle. For the same triangle, find and .
Relative to , the opposite side is and the adjacent side is , with the same hypotenuse:
Notice what happened. and . The two acute angles of a right triangle swap the roles of opposite and adjacent, which is why the sine of one equals the cosine of the other — and since , that is the beginning of the complementary-angle relations you meet in Part 2.
The mistake this example exists to prevent. "Opposite" and "adjacent" are not fixed labels on the triangle — they depend entirely on which angle you are working with. **The same side is opposite and adjacent to . A student who labels the triangle once and uses those labels for both angles will get half the answers wrong, and the error is invisible in the working.
Why the hypotenuse is special.** It is the longest side, so both and are always **less than ** for an acute angle, while and are always **greater than . And can be anything positive**, since neither of its sides is the hypotenuse. Those bounds are worth remembering as a first check on any answer.
How do you find the other five ratios when only one is given?
Read the given ratio as two sides of a right triangle, use the Pythagoras theorem to find the third, and then write all six ratios from the completed triangle.
The method is always the same three steps.
- Step 1 — from the given ratio, assign lengths to the two sides it names
- Step 2 — find the third side by Pythagoras
- Step 3 — read off whichever ratios are wanted
Worked example 1. If , find all the other trigonometric ratios of .
Step 1. Since , take the opposite side as and the adjacent side as for some positive . **Using rather than fixed numbers is what makes this rigorous — the ratio fixes the shape, not the size.
Step 2.** The hypotenuse is
Step 3. Now every ratio follows, and the cancels from each:
Check: . Correct, and as given.
**The cancelling is the point of the whole method. It shows that the answer does not depend on the size you chose — which is exactly the similarity fact the chapter opened with.
Worked example 2.** If , find and .
The opposite side is and the hypotenuse is , so the adjacent side is
Hence
Notice the subtraction. When the given ratio involves the hypotenuse, the third side comes from subtracting inside the square root, not adding. **Adding here would give and every answer would be wrong — so the first question to ask is always whether the hypotenuse is one of the two sides you were given.
Worked example 3.** If , find and .
Since , the hypotenuse is and the adjacent side is . The opposite side is
so
Check: , and . Correct.
Worked example 4 — a value that is impossible. Can for some acute angle ?
No. The sine of an acute angle is the opposite side over the hypotenuse, and the hypotenuse is the longest side of a right triangle, so the ratio must be less than . A value of would need the opposite side to exceed the hypotenuse, which is impossible.
**The same reasoning rules out **, and it permits and , since those two are always greater than . Checking a given value against these bounds before starting is worth ten seconds — it turns an unanswerable question into a one-line justification.
Which three ratios are bounded, and which are not.
- ** and ** lie strictly between and for an acute angle
- ** and ** are greater than
- ** and ** can take any positive value
The method is always the same three steps.
- Step 1 — from the given ratio, assign lengths to the two sides it names
- Step 2 — find the third side by Pythagoras
- Step 3 — read off whichever ratios are wanted
Worked example 1. If , find all the other trigonometric ratios of .
Step 1. Since , take the opposite side as and the adjacent side as for some positive . **Using rather than fixed numbers is what makes this rigorous — the ratio fixes the shape, not the size.
Step 2.** The hypotenuse is
Step 3. Now every ratio follows, and the cancels from each:
Check: . Correct, and as given.
**The cancelling is the point of the whole method. It shows that the answer does not depend on the size you chose — which is exactly the similarity fact the chapter opened with.
Worked example 2.** If , find and .
The opposite side is and the hypotenuse is , so the adjacent side is
Hence
Notice the subtraction. When the given ratio involves the hypotenuse, the third side comes from subtracting inside the square root, not adding. **Adding here would give and every answer would be wrong — so the first question to ask is always whether the hypotenuse is one of the two sides you were given.
Worked example 3.** If , find and .
Since , the hypotenuse is and the adjacent side is . The opposite side is
so
Check: , and . Correct.
Worked example 4 — a value that is impossible. Can for some acute angle ?
No. The sine of an acute angle is the opposite side over the hypotenuse, and the hypotenuse is the longest side of a right triangle, so the ratio must be less than . A value of would need the opposite side to exceed the hypotenuse, which is impossible.
**The same reasoning rules out **, and it permits and , since those two are always greater than . Checking a given value against these bounds before starting is worth ten seconds — it turns an unanswerable question into a one-line justification.
Which three ratios are bounded, and which are not.
- ** and ** lie strictly between and for an acute angle
- ** and ** are greater than
- ** and ** can take any positive value
What are the ratios of the standard angles and how do you use them?
Five angles have exact values you are expected to know without looking them up. Reading them across the table below is the fastest route into almost every numerical question in this chapter.
The sines, in order of angle:
The cosines are the same list read backwards:
The tangents come from dividing:
**Why is not defined.** It equals , and division by zero has no value. The correct phrase is "not defined", never "infinity" and never "undefined value" — and for the same reason , and are all not defined.
The one pattern that makes the table memorable. Write the sines as
**The numerators are the square roots of in order**, and every denominator is . Those are the same five values as above — being and being . The cosines are the same pattern in reverse, so one line reconstructs the whole table.
Worked example 1. Evaluate .
A striking answer, and not a coincidence. The expression is the expansion of , a formula you will meet in Class 11 — so the arithmetic has a reason behind it, and that agreement is itself a check.
Worked example 2. Evaluate .
The last two terms cancelled, because and are the same number. Spotting that saves the arithmetic entirely, and questions are often built so that such a cancellation is available.
Worked example 3 — a fraction with surds. Evaluate .
Substituting the values, with and :
Rationalising by multiplying top and bottom by , and using :
Check numerically. The answer is . Evaluating the original expression directly gives . The two agree, which is exactly the check to run whenever a surd simplification has several steps.
Worked example 4 — a longer expression. Evaluate .
The denominator is , so only the numerator matters:
Taking a common denominator of :
**Notice the denominator was before any work was done.** for every angle, so recognising it saves a calculation — and it is the identity Part 2 is built on.
Worked example 5 — a double angle in disguise. Evaluate .
**And is ** — which is , another Class 11 formula appearing early. Answers that turn out to be standard values are a strong sign the working is right.
The sines, in order of angle:
The cosines are the same list read backwards:
The tangents come from dividing:
**Why is not defined.** It equals , and division by zero has no value. The correct phrase is "not defined", never "infinity" and never "undefined value" — and for the same reason , and are all not defined.
The one pattern that makes the table memorable. Write the sines as
**The numerators are the square roots of in order**, and every denominator is . Those are the same five values as above — being and being . The cosines are the same pattern in reverse, so one line reconstructs the whole table.
Worked example 1. Evaluate .
A striking answer, and not a coincidence. The expression is the expansion of , a formula you will meet in Class 11 — so the arithmetic has a reason behind it, and that agreement is itself a check.
Worked example 2. Evaluate .
The last two terms cancelled, because and are the same number. Spotting that saves the arithmetic entirely, and questions are often built so that such a cancellation is available.
Worked example 3 — a fraction with surds. Evaluate .
Substituting the values, with and :
Rationalising by multiplying top and bottom by , and using :
Check numerically. The answer is . Evaluating the original expression directly gives . The two agree, which is exactly the check to run whenever a surd simplification has several steps.
Worked example 4 — a longer expression. Evaluate .
The denominator is , so only the numerator matters:
Taking a common denominator of :
**Notice the denominator was before any work was done.** for every angle, so recognising it saves a calculation — and it is the identity Part 2 is built on.
Worked example 5 — a double angle in disguise. Evaluate .
**And is ** — which is , another Class 11 formula appearing early. Answers that turn out to be standard values are a strong sign the working is right.
How do you solve for an unknown angle and judge a true-or-false claim?
Isolate the ratio, match it against the standard table, and then solve the small linear equation that remains for the angle.
Worked example 1. Solve for the acute angle .
Step 1 — isolate the ratio:
Step 2 — recognise the value. From the table, , so
**Step 3 — solve for :**
Check: . Correct.
The step students skip is the third one. Having found it is tempting to stop and write . **The angle you read off the table is , not **, and dividing it by the multiplier is a separate step. Write the multiplier down explicitly — ", so " — rather than carrying it in your head.
Worked example 2. Solve .
Check: and . Correct.
And here the substitution check does catch the dropped division. Had you answered , then would be , where the cosine is , giving instead of . Substituting into the original equation, not the rearranged one, exposes it at once — so make the substitution with the multiplier restored, exactly as the question wrote it.
Worked example 3. Solve and for acute angles and , given that .
From the table, and , so
Adding: , so . Subtracting: , so .
Check: and , and . All three conditions hold.
Notice that the trigonometry ended after one line. Once the angles were read off the table, the rest was a pair of linear equations — which is why this chapter sits so comfortably beside the earlier chapter on simultaneous equations.
Worked example 4 — finding an angle from a triangle. In , right-angled at , cm and . Find and .
Relative to , the side is opposite and is adjacent, so
and
Check by Pythagoras: . Correct.
Now the true-or-false questions, which test understanding rather than calculation.
Claim 1: for all angles.
False. Take and . The left side is . The right side is . Not equal, and a single counterexample settles a universal claim.
Claim 2: increases as increases, for .
True. From the table, , , and — steadily increasing. The reason is geometric: as the angle opens, the opposite side grows while the adjacent side shrinks, so the ratio rises on both counts.
Claim 3: is the product of and .
False. "" is the name of an operation, not a quantity. It has no meaning without an angle attached, so there is nothing for it to be a product with.
Claim 4: for some angle .
False, as shown earlier: the sine of an acute angle cannot exceed , because the hypotenuse is the longest side.
Claim 5: decreases as increases from to .
True. The values run , , , , — steadily falling. And it must, because , so as rises the complementary angle falls.
Worked example 1. Solve for the acute angle .
Step 1 — isolate the ratio:
Step 2 — recognise the value. From the table, , so
**Step 3 — solve for :**
Check: . Correct.
The step students skip is the third one. Having found it is tempting to stop and write . **The angle you read off the table is , not **, and dividing it by the multiplier is a separate step. Write the multiplier down explicitly — ", so " — rather than carrying it in your head.
Worked example 2. Solve .
Check: and . Correct.
And here the substitution check does catch the dropped division. Had you answered , then would be , where the cosine is , giving instead of . Substituting into the original equation, not the rearranged one, exposes it at once — so make the substitution with the multiplier restored, exactly as the question wrote it.
Worked example 3. Solve and for acute angles and , given that .
From the table, and , so
Adding: , so . Subtracting: , so .
Check: and , and . All three conditions hold.
Notice that the trigonometry ended after one line. Once the angles were read off the table, the rest was a pair of linear equations — which is why this chapter sits so comfortably beside the earlier chapter on simultaneous equations.
Worked example 4 — finding an angle from a triangle. In , right-angled at , cm and . Find and .
Relative to , the side is opposite and is adjacent, so
and
Check by Pythagoras: . Correct.
Now the true-or-false questions, which test understanding rather than calculation.
Claim 1: for all angles.
False. Take and . The left side is . The right side is . Not equal, and a single counterexample settles a universal claim.
Claim 2: increases as increases, for .
True. From the table, , , and — steadily increasing. The reason is geometric: as the angle opens, the opposite side grows while the adjacent side shrinks, so the ratio rises on both counts.
Claim 3: is the product of and .
False. "" is the name of an operation, not a quantity. It has no meaning without an angle attached, so there is nothing for it to be a product with.
Claim 4: for some angle .
False, as shown earlier: the sine of an acute angle cannot exceed , because the hypotenuse is the longest side.
Claim 5: decreases as increases from to .
True. The values run , , , , — steadily falling. And it must, because , so as rises the complementary angle falls.
Exam tip
Which habits keep a trigonometry answer safe?
Draw the triangle and label the sides relative to the angle you are actually working with. Nearly every error in this chapter is a labelling error, not an arithmetic one.
- Re-label for each angle. The side opposite is adjacent to , and using one set of labels for both angles is the classic mistake
- **Use when a ratio is given**, taking the sides as and ; the cancels and shows the answer is size-independent
- Ask whether the hypotenuse is one of the given sides. If it is, the third side comes from subtracting inside the square root
- **Check with — it is free and it catches almost everything
- Check with as a second, independent test
- Know the bounds**: and lie between and , and exceed , and and are unbounded
- **Rebuild the standard table from to rather than trusting memory under pressure
- Write "not defined"** for , , and
- Solve fully for the angle. From the answer is
- Substitute back into the original equation, not the rearranged one
- Rationalise surd answers and verify them with a quick decimal comparison
The misconception to name. , and are not quantities that can be cancelled or split. ** is not **, and is not . Each of these is a single operation applied to a single angle, and treating the name as a multiplier is the deepest error a student can make here — it survives into Class 11 and quietly ruins identity proofs.
A second trap. Giving where was asked for. The equation is solved for the angle inside the ratio, and then that has to be solved for the angle in the question — two steps, and the second is the one that gets dropped when time is short.
- Re-label for each angle. The side opposite is adjacent to , and using one set of labels for both angles is the classic mistake
- **Use when a ratio is given**, taking the sides as and ; the cancels and shows the answer is size-independent
- Ask whether the hypotenuse is one of the given sides. If it is, the third side comes from subtracting inside the square root
- **Check with — it is free and it catches almost everything
- Check with as a second, independent test
- Know the bounds**: and lie between and , and exceed , and and are unbounded
- **Rebuild the standard table from to rather than trusting memory under pressure
- Write "not defined"** for , , and
- Solve fully for the angle. From the answer is
- Substitute back into the original equation, not the rearranged one
- Rationalise surd answers and verify them with a quick decimal comparison
The misconception to name. , and are not quantities that can be cancelled or split. ** is not **, and is not . Each of these is a single operation applied to a single angle, and treating the name as a multiplier is the deepest error a student can make here — it survives into Class 11 and quietly ruins identity proofs.
A second trap. Giving where was asked for. The equation is solved for the angle inside the ratio, and then that has to be solved for the angle in the question — two steps, and the second is the one that gets dropped when time is short.
Did you know
Why do exactly three angles have such clean values?
The values for , and are not arbitrary. Each comes from cutting a very simple figure in half, and you can derive all three in about a minute.
**For , cut a square along its diagonal.** The result is a right triangle with two equal sides — say both of length — and a hypotenuse of . Both acute angles are , so
And it is obvious from the figure why the sine and cosine are equal — the two legs are the same length, so it makes no difference which one you call opposite.
**For and , cut an equilateral triangle down the middle.** Take one of side . The altitude splits the base into two halves of length and bisects the angle at the top into two angles. Its own length is
**So the half-triangle has sides , and **, with the angle at the top and the angle at the base. Reading the ratios off it:
- **From the angle**, the opposite side is and the adjacent is , giving , and
- **From the angle**, opposite and adjacent swap, giving , and
That is the whole table, from a square and an equilateral triangle. And it explains why the sines of and are each other's cosines: they are the two acute angles of the same triangle.
**The values for and come from a different argument, because no triangle has a zero angle. Instead, imagine the angle shrinking toward zero:
- The opposite side shrinks to nothing** while the adjacent side approaches the hypotenuse, so and
- **As the angle opens toward ** the opposite side approaches the hypotenuse and the adjacent side shrinks to nothing, so and
**And that is exactly why is not defined — the adjacent side has shrunk to zero, and you cannot divide by it. The ratio does not grow to some enormous number and stop; it has no value at all, which is why the phrase matters.
One last observation about why and and keep appearing. They are the only lengths a square and an equilateral triangle can produce from a unit side. No other angle between and comes from halving such a simple figure**, which is why has no neat form and why these five angles are the ones every examination uses.
**For , cut a square along its diagonal.** The result is a right triangle with two equal sides — say both of length — and a hypotenuse of . Both acute angles are , so
And it is obvious from the figure why the sine and cosine are equal — the two legs are the same length, so it makes no difference which one you call opposite.
**For and , cut an equilateral triangle down the middle.** Take one of side . The altitude splits the base into two halves of length and bisects the angle at the top into two angles. Its own length is
**So the half-triangle has sides , and **, with the angle at the top and the angle at the base. Reading the ratios off it:
- **From the angle**, the opposite side is and the adjacent is , giving , and
- **From the angle**, opposite and adjacent swap, giving , and
That is the whole table, from a square and an equilateral triangle. And it explains why the sines of and are each other's cosines: they are the two acute angles of the same triangle.
**The values for and come from a different argument, because no triangle has a zero angle. Instead, imagine the angle shrinking toward zero:
- The opposite side shrinks to nothing** while the adjacent side approaches the hypotenuse, so and
- **As the angle opens toward ** the opposite side approaches the hypotenuse and the adjacent side shrinks to nothing, so and
**And that is exactly why is not defined — the adjacent side has shrunk to zero, and you cannot divide by it. The ratio does not grow to some enormous number and stop; it has no value at all, which is why the phrase matters.
One last observation about why and and keep appearing. They are the only lengths a square and an equilateral triangle can produce from a unit side. No other angle between and comes from halving such a simple figure**, which is why has no neat form and why these five angles are the ones every examination uses.
Exam relevance
How does trigonometry at this level prepare you for JEE and NEET?
This is foundation work for Class 11 Trigonometric Functions and for the vector and oscillation topics in Physics, so it feeds JEE Main, JEE Advanced and NEET alike.
Where the six ratios lead. Class 11 redefines them for any angle, positive or negative, using a circle rather than a triangle, and adds the sign rules for the four quadrants. The definitions you learn here become the first-quadrant case of that general system, and the numerical values for the five standard angles are carried over unchanged and used constantly.
Where the standard-angle table leads. It grows to include , , , and , all derived from the compound-angle and half-angle formulas. **The two accidental appearances in this page — , and — are those formulas showing up early, and recognising them now makes Class 11 feel like a continuation rather than a fresh start.
Where the find-all-six-from-one method leads. It is used throughout Class 11 with the added complication of deciding the sign from the quadrant, and it is the standard opening move in JEE questions that supply one ratio and ask for an expression in others. The habit of introducing and letting it cancel survives unchanged.
Where the identity check leads.** is the first of the three Pythagorean identities, and it is the single most used relation in all of trigonometry. **Recognising it inside a longer expression, as in the example, is exactly the skill JEE tests — usually buried two or three steps deep.
Where the ratios lead in Physics.** Resolving a force or a velocity into components is and , and it appears in the very first mechanics chapter of Class 11. NEET Physics uses it in every problem on inclined planes, projectiles and optics, and a student slow with the standard values is slow in all of them.
Question types to expect. At this level: compute the six ratios, find the rest from one, evaluate a standard-angle expression, and solve for an acute angle. In competitive papers: quadrant signs, compound and multiple angles, trigonometric equations with general solutions, and component resolution in Physics.
The single trap that costs marks. Treating as a multiplier. **** — the counterexample , settles it — and the same illegitimate splitting in Class 11 identity proofs produces answers that look plausible and are wrong throughout.
A second trap. Stopping at the inner angle. **From you get and then — and in Class 11, where the general solution adds a whole family of angles, dropping that division causes every member of the family to be wrong.
Board versus competitive emphasis. The CBSE paper marks the labelled triangle, the Pythagoras step, the ratios and the substituted values; a competitive paper marks the final number, often reached from a value in a non-standard quadrant. The transferable habit is verifying with ** — it costs one line, it uses only what you already have, and it is the check that keeps working no matter how far the subject goes.
Where the six ratios lead. Class 11 redefines them for any angle, positive or negative, using a circle rather than a triangle, and adds the sign rules for the four quadrants. The definitions you learn here become the first-quadrant case of that general system, and the numerical values for the five standard angles are carried over unchanged and used constantly.
Where the standard-angle table leads. It grows to include , , , and , all derived from the compound-angle and half-angle formulas. **The two accidental appearances in this page — , and — are those formulas showing up early, and recognising them now makes Class 11 feel like a continuation rather than a fresh start.
Where the find-all-six-from-one method leads. It is used throughout Class 11 with the added complication of deciding the sign from the quadrant, and it is the standard opening move in JEE questions that supply one ratio and ask for an expression in others. The habit of introducing and letting it cancel survives unchanged.
Where the identity check leads.** is the first of the three Pythagorean identities, and it is the single most used relation in all of trigonometry. **Recognising it inside a longer expression, as in the example, is exactly the skill JEE tests — usually buried two or three steps deep.
Where the ratios lead in Physics.** Resolving a force or a velocity into components is and , and it appears in the very first mechanics chapter of Class 11. NEET Physics uses it in every problem on inclined planes, projectiles and optics, and a student slow with the standard values is slow in all of them.
Question types to expect. At this level: compute the six ratios, find the rest from one, evaluate a standard-angle expression, and solve for an acute angle. In competitive papers: quadrant signs, compound and multiple angles, trigonometric equations with general solutions, and component resolution in Physics.
The single trap that costs marks. Treating as a multiplier. **** — the counterexample , settles it — and the same illegitimate splitting in Class 11 identity proofs produces answers that look plausible and are wrong throughout.
A second trap. Stopping at the inner angle. **From you get and then — and in Class 11, where the general solution adds a whole family of angles, dropping that division causes every member of the family to be wrong.
Board versus competitive emphasis. The CBSE paper marks the labelled triangle, the Pythagoras step, the ratios and the substituted values; a competitive paper marks the final number, often reached from a value in a non-standard quadrant. The transferable habit is verifying with ** — it costs one line, it uses only what you already have, and it is the check that keeps working no matter how far the subject goes.
Key takeaways
What must you be able to do from this part?
Six ratios, five angles and one identity that checks everything.
- A ratio of two sides of a right triangle depends only on the angle, because all such triangles with that angle are similar
- **, , , and cosec, sec and cot are their reciprocals
- and
- Opposite and adjacent are relative to the angle**, so re-label for each angle — the side opposite is adjacent to
- **With and about a right angle at **, the hypotenuse is , giving and
- ** and **, since and are complementary
- **From one ratio, take the two sides as multiples of **, find the third by Pythagoras, and let cancel — gives and
- If the hypotenuse is given, subtract inside the root — gives an adjacent side of
- Bounds: and lie between and , cosec and sec exceed , and and are unbounded — so is impossible
- **The sines of to are to , and the cosines are the same list reversed
- , , and are not defined
- **, and
- ****, and
- ** for every angle** — spot it and a denominator may vanish, as in the example
- Solve fully for the angle: gives , and gives
- ** with ** gives and
- ** — the name is an operation, not a multiplier
- and come from halving an equilateral triangle; from halving a square**
The fastest self-test needs no table at all. Draw an equilateral triangle of side , cut it in half, and read off all six ratios of both and from the figure — if you can do that in under a minute, the table will never desert you in an examination.
- A ratio of two sides of a right triangle depends only on the angle, because all such triangles with that angle are similar
- **, , , and cosec, sec and cot are their reciprocals
- and
- Opposite and adjacent are relative to the angle**, so re-label for each angle — the side opposite is adjacent to
- **With and about a right angle at **, the hypotenuse is , giving and
- ** and **, since and are complementary
- **From one ratio, take the two sides as multiples of **, find the third by Pythagoras, and let cancel — gives and
- If the hypotenuse is given, subtract inside the root — gives an adjacent side of
- Bounds: and lie between and , cosec and sec exceed , and and are unbounded — so is impossible
- **The sines of to are to , and the cosines are the same list reversed
- , , and are not defined
- **, and
- ****, and
- ** for every angle** — spot it and a denominator may vanish, as in the example
- Solve fully for the angle: gives , and gives
- ** with ** gives and
- ** — the name is an operation, not a multiplier
- and come from halving an equilateral triangle; from halving a square**
The fastest self-test needs no table at all. Draw an equilateral triangle of side , cut it in half, and read off all six ratios of both and from the figure — if you can do that in under a minute, the table will never desert you in an examination.