The Longest Side Always Faces the Largest Angle
Rank a triangle's angles from its sides and its sides from its angles, test whether three lengths can form a triangle at all, prove the perpendicular is the shortest route to a line, and order the parts of a figure.
Which angle of a triangle is the biggest if you only know the sides?
Part 1 dealt with equal sides and equal angles. Part 2 asks the same question about unequal ones, and the answer is just as tidy.
In words: the angle opposite the greater side is greater. And the converse holds too — the side opposite the greater angle is greater.
You can see why in one picture. Keep two sides of a triangle fixed and open the angle between them: the third side gets longer. Open it further and it gets longer still. The opening and the side facing it grow together, so the largest opening faces the longest side.
Worked example. In , cm, cm and cm. Which angle is largest and which is smallest?
First pair each side with the angle opposite it — the angle at the vertex the side does not touch:
- cm is opposite
- cm is opposite
- cm is opposite
Since , the angles rank the same way: . **So the largest angle is and the smallest is ** — and no angle was measured.
This page covers the second part of the ICSE Class 9 Mathematics chapter on isosceles triangles and inequalities: ranking sides and angles, testing whether three lengths form a triangle, proving that the perpendicular is the shortest distance to a line, and arranging the parts of a figure in order.
In words: the angle opposite the greater side is greater. And the converse holds too — the side opposite the greater angle is greater.
You can see why in one picture. Keep two sides of a triangle fixed and open the angle between them: the third side gets longer. Open it further and it gets longer still. The opening and the side facing it grow together, so the largest opening faces the longest side.
Worked example. In , cm, cm and cm. Which angle is largest and which is smallest?
First pair each side with the angle opposite it — the angle at the vertex the side does not touch:
- cm is opposite
- cm is opposite
- cm is opposite
Since , the angles rank the same way: . **So the largest angle is and the smallest is ** — and no angle was measured.
This page covers the second part of the ICSE Class 9 Mathematics chapter on isosceles triangles and inequalities: ranking sides and angles, testing whether three lengths form a triangle, proving that the perpendicular is the shortest distance to a line, and arranging the parts of a figure in order.
How do you find the longest side when you are given the angles?
Rank the angles, then read off the sides opposite them in the same order. The converse theorem lets you run the previous section backwards.
Worked example 1. In , and . Arrange the sides in ascending order.
First find the third angle:
Now rank: . The sides opposite these are , and , so
Worked example 2 — a right-angled triangle. In , . Which side is longest?
The other two angles must add to , so each is less than . The right angle is therefore the largest angle, and the side opposite it — the hypotenuse — is the longest side.
This is worth stating as a permanent fact: the hypotenuse is always the longest side of a right-angled triangle, and now you know it is not a special rule but the ranking theorem applied once.
Worked example 3 — an obtuse triangle. If in , what can you say about ?
The other two angles add to , so both are less than . So is the largest and , the side opposite it, is the longest side.
**A triangle can have at most one angle of or more, so there is never any ambiguity about which side is longest in a right-angled or obtuse triangle.
One misconception to clear. The theorem compares sides within one triangle**. A angle in a large triangle can face a much longer side than an angle in a small one. The ranking is internal, and comparing parts across two different triangles needs congruence or similarity, not this theorem.
Worked example 1. In , and . Arrange the sides in ascending order.
First find the third angle:
Now rank: . The sides opposite these are , and , so
Worked example 2 — a right-angled triangle. In , . Which side is longest?
The other two angles must add to , so each is less than . The right angle is therefore the largest angle, and the side opposite it — the hypotenuse — is the longest side.
This is worth stating as a permanent fact: the hypotenuse is always the longest side of a right-angled triangle, and now you know it is not a special rule but the ranking theorem applied once.
Worked example 3 — an obtuse triangle. If in , what can you say about ?
The other two angles add to , so both are less than . So is the largest and , the side opposite it, is the longest side.
**A triangle can have at most one angle of or more, so there is never any ambiguity about which side is longest in a right-angled or obtuse triangle.
One misconception to clear. The theorem compares sides within one triangle**. A angle in a large triangle can face a much longer side than an angle in a small one. The ranking is internal, and comparing parts across two different triangles needs congruence or similarity, not this theorem.
Can any three lengths make a triangle?
No. The sum of any two sides must be greater than the third side — and that must hold for all three pairs.
Why it must be true follows from the section above. Suppose you try to build a triangle with sides , and cm. Lay the cm side down, open a cm arm from one end and an cm arm from the other. The two arms together reach only cm — they cannot meet. The two shorter sides have to be long enough to bridge the longest one.
Worked example 1 — the test in practice. Which of these can be the sides of a triangle?
- cm, cm, cm: , , . Yes
- cm, cm, cm: , which is not greater than . No
- cm, cm, cm: , and the other two pairs are clearly fine. Yes
- cm, cm, cm: , which is not greater than . No
The shortcut: you only need to check the two smallest against the largest. If that pair passes, the other two pairs pass automatically, because each of them already contains the largest side.
The fourth case is the interesting one. With exactly, the two short sides lie flat along the long one and the three points fall on a single straight line. There is no triangle — only a degenerate figure with zero area. The inequality is strict for a reason, and questions test that boundary deliberately.
Worked example 2 — the range of the third side. Two sides of a triangle are cm and cm. Between what limits must the third side lie?
The upper limit comes from the sum:
The lower limit comes from the other two inequalities. From we get , which is the difference of the given sides. So
The rule in general: the third side lies strictly between the difference and the sum of the other two. Both limits are excluded, since either would flatten the triangle.
Check it: cm gives , so it works; cm gives , exactly flat, so it does not. The boundary values are the answers to reject, and writing the inequality with strict signs is what shows you know that.
Why it must be true follows from the section above. Suppose you try to build a triangle with sides , and cm. Lay the cm side down, open a cm arm from one end and an cm arm from the other. The two arms together reach only cm — they cannot meet. The two shorter sides have to be long enough to bridge the longest one.
Worked example 1 — the test in practice. Which of these can be the sides of a triangle?
- cm, cm, cm: , , . Yes
- cm, cm, cm: , which is not greater than . No
- cm, cm, cm: , and the other two pairs are clearly fine. Yes
- cm, cm, cm: , which is not greater than . No
The shortcut: you only need to check the two smallest against the largest. If that pair passes, the other two pairs pass automatically, because each of them already contains the largest side.
The fourth case is the interesting one. With exactly, the two short sides lie flat along the long one and the three points fall on a single straight line. There is no triangle — only a degenerate figure with zero area. The inequality is strict for a reason, and questions test that boundary deliberately.
Worked example 2 — the range of the third side. Two sides of a triangle are cm and cm. Between what limits must the third side lie?
The upper limit comes from the sum:
The lower limit comes from the other two inequalities. From we get , which is the difference of the given sides. So
The rule in general: the third side lies strictly between the difference and the sum of the other two. Both limits are excluded, since either would flatten the triangle.
Check it: cm gives , so it works; cm gives , exactly flat, so it does not. The boundary values are the answers to reject, and writing the inequality with strict signs is what shows you know that.
Why is the perpendicular the shortest route from a point to a line?
Because the perpendicular foot is the only point where the angle at the line is a right angle, and the right angle forces the opposite side to be the longest — which means every other route is longer than the perpendicular.
Here is the proof, and it uses nothing but the ranking theorem.
Let be a point not on the line , let be perpendicular to with on the line, and let be any other point of .
In :
- (given, )
- so , which makes
- therefore
By the ranking theorem the side opposite the greater angle is greater, so
Since was any point of the line other than , the perpendicular is shorter than every other segment from to .
**That is why distance from a point to a line means the perpendicular distance — the phrase would be meaningless otherwise, since there are infinitely many segments to choose from.
Worked example.** A point is cm from a line , measured perpendicularly, and is a point of that is cm from the foot . How far is ?
is right-angled at , so
And , as the theorem promised. **Move further along the line and grows without limit, but it never falls below cm.
One consequence used constantly later. Of all the chords through a point inside a circle, and of all the segments from a centre to a line, the perpendicular is the one the geometry singles out. Every formula you meet later that says distance is quietly saying perpendicular distance** — including the area of a triangle as half base times height, where the height must be measured perpendicular to the base and not along a slanting side.
Here is the proof, and it uses nothing but the ranking theorem.
Let be a point not on the line , let be perpendicular to with on the line, and let be any other point of .
In :
- (given, )
- so , which makes
- therefore
By the ranking theorem the side opposite the greater angle is greater, so
Since was any point of the line other than , the perpendicular is shorter than every other segment from to .
**That is why distance from a point to a line means the perpendicular distance — the phrase would be meaningless otherwise, since there are infinitely many segments to choose from.
Worked example.** A point is cm from a line , measured perpendicularly, and is a point of that is cm from the foot . How far is ?
is right-angled at , so
And , as the theorem promised. **Move further along the line and grows without limit, but it never falls below cm.
One consequence used constantly later. Of all the chords through a point inside a circle, and of all the segments from a centre to a line, the perpendicular is the one the geometry singles out. Every formula you meet later that says distance is quietly saying perpendicular distance** — including the area of a triangle as half base times height, where the height must be measured perpendicular to the base and not along a slanting side.
How do you arrange the sides or angles of a whole figure in order?
Work one triangle at a time, write the inequality each gives you, and only then combine them through the part they share.
Worked example 1 — one triangle. In , and . Arrange the sides in descending order.
, so , and the sides opposite give
Worked example 2 — two triangles sharing a side. In the figure, and are triangles on opposite sides of , with cm, cm, cm and cm. Compare with , and with .
In : , so the angle opposite is greater, giving .
In : , so .
Now add the two results. , and those sums are the full angles of the quadrilateral at and at :
That addition step is the whole technique for figure questions. Each triangle contributes one inequality about its own angles, and the shared side lets you add them into a statement about the figure.
Worked example 3 — the general quadrilateral result. In quadrilateral , is the shortest side and is the longest. Show that .
Join . In , (since is shortest), so . In , (since is longest), so . Adding gives .
The diagonal was the construction. Just as Part 1 needed the apex bisector, inequality riders need a diagonal or a perpendicular — the figure as given rarely contains the triangles you need.
And one restriction on adding inequalities. You may add two inequalities that point the same way, as above. You may not subtract them: from and it does not follow that . Adding is safe, subtracting is not, and a rider that seems to need subtraction almost always needs a different pair of triangles instead.
Worked example 1 — one triangle. In , and . Arrange the sides in descending order.
, so , and the sides opposite give
Worked example 2 — two triangles sharing a side. In the figure, and are triangles on opposite sides of , with cm, cm, cm and cm. Compare with , and with .
In : , so the angle opposite is greater, giving .
In : , so .
Now add the two results. , and those sums are the full angles of the quadrilateral at and at :
That addition step is the whole technique for figure questions. Each triangle contributes one inequality about its own angles, and the shared side lets you add them into a statement about the figure.
Worked example 3 — the general quadrilateral result. In quadrilateral , is the shortest side and is the longest. Show that .
Join . In , (since is shortest), so . In , (since is longest), so . Adding gives .
The diagonal was the construction. Just as Part 1 needed the apex bisector, inequality riders need a diagonal or a perpendicular — the figure as given rarely contains the triangles you need.
And one restriction on adding inequalities. You may add two inequalities that point the same way, as above. You may not subtract them: from and it does not follow that . Adding is safe, subtracting is not, and a rider that seems to need subtraction almost always needs a different pair of triangles instead.
Exam tip
What layout keeps an inequality proof from losing marks?
Name the triangle before every inequality, and give the reason in brackets. An inequality with no triangle attached is meaningless, because these theorems only compare parts of the same triangle.
- **Write *In * at the start of each line. Two-triangle riders live or die on keeping the two arguments separate before they are added
- Quote the reason as content**: (angle opposite the greater side) or (side opposite the greater angle). Do not write by theorem
- Pair each side with the opposite vertex in writing before ranking anything. faces , and the vertex letters make this easy to get wrong at speed
- For the three-length test, check only the two smallest against the largest and say so. It is a complete answer and it saves two lines
- Use strict inequality signs everywhere. , never . The equality case is a straight line, not a triangle, and the mark is for knowing that
- For a range question, give both limits with units: . A single-sided answer is a half answer
- Add inequalities, never subtract them, and write the addition as its own line
The trap that looks like arithmetic but is not. In a range question, the difference gives the lower limit and the sum gives the upper limit — students frequently produce by subtracting nothing at all. Test your lower limit by putting the smallest allowed value into , and the error shows up at once.
- **Write *In * at the start of each line. Two-triangle riders live or die on keeping the two arguments separate before they are added
- Quote the reason as content**: (angle opposite the greater side) or (side opposite the greater angle). Do not write by theorem
- Pair each side with the opposite vertex in writing before ranking anything. faces , and the vertex letters make this easy to get wrong at speed
- For the three-length test, check only the two smallest against the largest and say so. It is a complete answer and it saves two lines
- Use strict inequality signs everywhere. , never . The equality case is a straight line, not a triangle, and the mark is for knowing that
- For a range question, give both limits with units: . A single-sided answer is a half answer
- Add inequalities, never subtract them, and write the addition as its own line
The trap that looks like arithmetic but is not. In a range question, the difference gives the lower limit and the sum gives the upper limit — students frequently produce by subtracting nothing at all. Test your lower limit by putting the smallest allowed value into , and the error shows up at once.
Did you know
Why does everyone cut across the maidan instead of walking round it?
There is a worn diagonal path across almost every park and college ground, cutting the corner that the paved footpaths go round. Nobody plans those paths; they appear because walkers are solving a geometry problem without noticing.
Going round two sides of a triangle covers ; cutting across covers . The triangle inequality says , so the diagonal is always shorter, for every possible triangle. The saving is not a special feature of one park — it is a theorem.
The inequality also tells you how much you save, or rather, how much you cannot. Since as well, the shortcut can never be shorter than the difference of the two sides. So a detour of m and m along two paths can be replaced by a diagonal somewhere between m and m — and only the actual angle at the corner decides where in that range it falls.
The same statement is why a stretched string is straight. A taut thread between two pins takes the shortest available route, and any third point it passed through would make the path a sum of two sides instead of one. The straightness of a stretched string is the triangle inequality made physical.
And it is why this result matters far beyond triangles. In later mathematics the triangle inequality becomes the defining property of distance itself: any sensible measure of distance, on a map, on a graph or between two vectors, must satisfy . You are meeting one of the few statements in school geometry that survives essentially unchanged into university mathematics.
Going round two sides of a triangle covers ; cutting across covers . The triangle inequality says , so the diagonal is always shorter, for every possible triangle. The saving is not a special feature of one park — it is a theorem.
The inequality also tells you how much you save, or rather, how much you cannot. Since as well, the shortcut can never be shorter than the difference of the two sides. So a detour of m and m along two paths can be replaced by a diagonal somewhere between m and m — and only the actual angle at the corner decides where in that range it falls.
The same statement is why a stretched string is straight. A taut thread between two pins takes the shortest available route, and any third point it passed through would make the path a sum of two sides instead of one. The straightness of a stretched string is the triangle inequality made physical.
And it is why this result matters far beyond triangles. In later mathematics the triangle inequality becomes the defining property of distance itself: any sensible measure of distance, on a map, on a graph or between two vectors, must satisfy . You are meeting one of the few statements in school geometry that survives essentially unchanged into university mathematics.
Exam relevance
How are triangle inequalities used in JEE-level questions?
This is foundation work with an unusually long reach, because the triangle inequality is a tool rather than a fact to recall.
Where it leads. In Class 11 the ranking theorem becomes quantitative in the Sine Rule — the side opposite a bigger angle is bigger because is constant — and the triangle inequality reappears in Vectors as , which is examined directly in JEE Main. In Complex Numbers the same statement appears as , one of the most frequently used inequalities in the whole syllabus.
Where the specific results survive. Perpendicular distance is the definition used in every Straight Lines question — the distance from a point to a line, the distance between parallel lines, the shortest distance from a centre to a chord — and the proof that it is shortest is the one you met above. In Physics, resolving a force perpendicular to a surface uses the same figure.
Question types to expect. At this level: rank the sides or angles, test three lengths, find the range of a third side. In competitive papers: the range of a parameter for which a triangle exists, and vector or modulus inequalities where the equality case has to be identified. Assertion-reason items like the degenerate case — three lengths with — because it looks like a triangle in a hurried reading.
The single trap that costs marks. Writing instead of . The equality case is collinear, so an answer that includes the endpoints of the range is wrong, and in JEE the difference between an open and a closed interval is the whole question.
Board versus competitive emphasis. ICSE asks you to prove and to arrange, so the named reasons carry marks. A competitive paper asks for the interval or the maximum value, so the premium is on knowing the strictness and the equality condition. Both reward the same habit: say exactly when the inequality becomes an equality.
Where it leads. In Class 11 the ranking theorem becomes quantitative in the Sine Rule — the side opposite a bigger angle is bigger because is constant — and the triangle inequality reappears in Vectors as , which is examined directly in JEE Main. In Complex Numbers the same statement appears as , one of the most frequently used inequalities in the whole syllabus.
Where the specific results survive. Perpendicular distance is the definition used in every Straight Lines question — the distance from a point to a line, the distance between parallel lines, the shortest distance from a centre to a chord — and the proof that it is shortest is the one you met above. In Physics, resolving a force perpendicular to a surface uses the same figure.
Question types to expect. At this level: rank the sides or angles, test three lengths, find the range of a third side. In competitive papers: the range of a parameter for which a triangle exists, and vector or modulus inequalities where the equality case has to be identified. Assertion-reason items like the degenerate case — three lengths with — because it looks like a triangle in a hurried reading.
The single trap that costs marks. Writing instead of . The equality case is collinear, so an answer that includes the endpoints of the range is wrong, and in JEE the difference between an open and a closed interval is the whole question.
Board versus competitive emphasis. ICSE asks you to prove and to arrange, so the named reasons carry marks. A competitive paper asks for the interval or the maximum value, so the premium is on knowing the strictness and the equality condition. Both reward the same habit: say exactly when the inequality becomes an equality.
Key takeaways
What should you be able to do with triangle inequalities before moving on?
Part 2 replaces equal with greater, and almost everything follows from a single theorem.
- The angle opposite the greater side is greater, and the converse — the side opposite the greater angle is greater
- Pair every side with the vertex it does not touch before ranking anything
- The hypotenuse is the longest side of a right-angled triangle, and the side facing an obtuse angle is longest in an obtuse one
- Three lengths form a triangle only if the sum of any two exceeds the third, and checking the two smallest against the largest is enough
- The third side lies strictly between the difference and the sum of the other two — both limits excluded
- Equality gives a degenerate straight line, not a triangle, so every sign is strict
- The perpendicular is the shortest segment from a point to a line, proved by the ranking theorem inside a right-angled triangle
- For figure questions, take one triangle at a time and add the inequalities — never subtract them, and expect to draw a diagonal
The quickest self-test is the range question: two sides of cm and cm, and the third side somewhere between. Write the limits down, then justify each one from a different inequality — and check whether you reached for the strict signs without being reminded.
- The angle opposite the greater side is greater, and the converse — the side opposite the greater angle is greater
- Pair every side with the vertex it does not touch before ranking anything
- The hypotenuse is the longest side of a right-angled triangle, and the side facing an obtuse angle is longest in an obtuse one
- Three lengths form a triangle only if the sum of any two exceeds the third, and checking the two smallest against the largest is enough
- The third side lies strictly between the difference and the sum of the other two — both limits excluded
- Equality gives a degenerate straight line, not a triangle, so every sign is strict
- The perpendicular is the shortest segment from a point to a line, proved by the ranking theorem inside a right-angled triangle
- For figure questions, take one triangle at a time and add the inequalities — never subtract them, and expect to draw a diagonal
The quickest self-test is the range question: two sides of cm and cm, and the third side somewhere between. Write the limits down, then justify each one from a different inequality — and check whether you reached for the strict signs without being reminded.