Two Measurements Are Enough to Solve a Whole Triangle
Find every unknown side and angle of a right-angled triangle from two given parts, use standard angles to compute a length, solve pole and shadow problems, and pick the right triangle out of a larger figure.
How little do you need to know before a right triangle is fully determined?
A triangle has six parts — three sides and three angles. In a right-angled triangle one of them is already known, since one angle is . So how many of the remaining five do you need?
Two, as long as they are not both angles.
That follows from what you already know. Two angles fix only the shape, not the size, so two angles are not enough. But one side and one acute angle is enough: the angle gives you every ratio, and the side turns a ratio into a length. And two sides is enough as well: the third comes from the Pythagoras theorem and the angles come from the ratios.
Solving a triangle means finding all the missing parts. Here is the whole method in three lines:
- The third angle comes from minus the known acute angle
- A missing side comes from whichever ratio connects it to a side you already have
- A missing angle comes from whichever ratio you can compute from two known sides
The only decision is which ratio to use, and the rule for that is simple: pick the one that contains the two things you know and the one thing you want.
This page covers the ICSE Class 9 Mathematics chapter on the solution of right triangles: solving a triangle completely, computing sides with standard angles, pole and shadow problems, and finding the right triangle hidden inside a larger figure.
Two, as long as they are not both angles.
That follows from what you already know. Two angles fix only the shape, not the size, so two angles are not enough. But one side and one acute angle is enough: the angle gives you every ratio, and the side turns a ratio into a length. And two sides is enough as well: the third comes from the Pythagoras theorem and the angles come from the ratios.
Solving a triangle means finding all the missing parts. Here is the whole method in three lines:
- The third angle comes from minus the known acute angle
- A missing side comes from whichever ratio connects it to a side you already have
- A missing angle comes from whichever ratio you can compute from two known sides
The only decision is which ratio to use, and the rule for that is simple: pick the one that contains the two things you know and the one thing you want.
This page covers the ICSE Class 9 Mathematics chapter on the solution of right triangles: solving a triangle completely, computing sides with standard angles, pole and shadow problems, and finding the right triangle hidden inside a larger figure.
How do you solve a right triangle completely from the given data?
Find the third angle first — it takes one subtraction — then use one ratio for each missing side.
Worked example 1 — a side and an angle. In , , and cm. Solve the triangle.
The third angle: .
is the hypotenuse, and is opposite , so use the sine:
is adjacent to , so use the cosine:
Check with the Pythagoras theorem: , as required.
Worked example 2 — two sides given. In , , cm and cm. Solve the triangle.
is adjacent to and is the hypotenuse, so
Then cm.
Check: , as required.
Worked example 3 — two equal legs. If both legs of a right-angled triangle are cm, solve it.
One check to run on every solved triangle. The longest side must face the largest angle. In example 1 the sides are , and , facing , and — in the same order, as the inequalities chapter requires. If your smallest side faces your largest angle, you have used the wrong ratio somewhere, and that check costs five seconds.
Worked example 1 — a side and an angle. In , , and cm. Solve the triangle.
The third angle: .
is the hypotenuse, and is opposite , so use the sine:
is adjacent to , so use the cosine:
Check with the Pythagoras theorem: , as required.
Worked example 2 — two sides given. In , , cm and cm. Solve the triangle.
is adjacent to and is the hypotenuse, so
Then cm.
Check: , as required.
Worked example 3 — two equal legs. If both legs of a right-angled triangle are cm, solve it.
One check to run on every solved triangle. The longest side must face the largest angle. In example 1 the sides are , and , facing , and — in the same order, as the inequalities chapter requires. If your smallest side faces your largest angle, you have used the wrong ratio somewhere, and that check costs five seconds.
How do you calculate an unknown side using a standard angle?
Write the ratio that connects the known side to the wanted side, substitute the standard value, and rearrange once.
The pattern is always the same three-line shape, and getting into the habit of writing all three lines is what protects the marks.
Worked example 1 — the wanted side on top. A tower is seen at an angle of elevation of from a point m from its foot on level ground. Find its height.
The height is opposite the angle and the m is adjacent to it, so use the tangent:
Worked example 2 — the wanted side underneath. A kite is flying at a height of m, and its string makes an angle of with the ground. Find the length of the string, assuming it is straight.
The height is opposite the angle and the string is the hypotenuse:
When the unknown is in the denominator, the answer comes out bigger, and that is worth noticing as a check: a string must be longer than the height it reaches.
Worked example 3 — choosing between two ratios. A ladder m long leans against a wall at to the ground. How high up the wall does it reach, and how far is its foot from the wall?
The height is opposite the angle, the distance from the wall is adjacent, and the ladder is the hypotenuse:
Check: , as required.
Worked example 4 — an angle of depression. From the top of a m tower, a boat is seen at an angle of depression of . How far is the boat from the foot of the tower?
The angle of depression from the top equals the angle of elevation from the boat, because they are alternate angles between the horizontal at the top and the horizontal ground. So
That equality of elevation and depression is worth stating as a step in your answer, because it is the step examiners look for. Depression is measured downwards from the horizontal, elevation upwards — and the two are equal for the same pair of points, which is why a question about depression can always be redrawn as one about elevation.
The pattern is always the same three-line shape, and getting into the habit of writing all three lines is what protects the marks.
Worked example 1 — the wanted side on top. A tower is seen at an angle of elevation of from a point m from its foot on level ground. Find its height.
The height is opposite the angle and the m is adjacent to it, so use the tangent:
Worked example 2 — the wanted side underneath. A kite is flying at a height of m, and its string makes an angle of with the ground. Find the length of the string, assuming it is straight.
The height is opposite the angle and the string is the hypotenuse:
When the unknown is in the denominator, the answer comes out bigger, and that is worth noticing as a check: a string must be longer than the height it reaches.
Worked example 3 — choosing between two ratios. A ladder m long leans against a wall at to the ground. How high up the wall does it reach, and how far is its foot from the wall?
The height is opposite the angle, the distance from the wall is adjacent, and the ladder is the hypotenuse:
Check: , as required.
Worked example 4 — an angle of depression. From the top of a m tower, a boat is seen at an angle of depression of . How far is the boat from the foot of the tower?
The angle of depression from the top equals the angle of elevation from the boat, because they are alternate angles between the horizontal at the top and the horizontal ground. So
That equality of elevation and depression is worth stating as a step in your answer, because it is the step examiners look for. Depression is measured downwards from the horizontal, elevation upwards — and the two are equal for the same pair of points, which is why a question about depression can always be redrawn as one about elevation.
How do you solve a pole and shadow problem?
The sun's elevation, the object's height and its shadow are the three parts of one right-angled triangle, and the tangent connects them.
Worked example 1. A pole casts a shadow m long when the sun's elevation is . Find the height of the pole.
Worked example 2 — the same pole later in the day. Find the shadow of that pole when the sun's elevation rises to .
**So the shadow shrank from m to m** while the sun climbed from to . The height never changed, of course — and that fixed height is what lets you use one calculation to feed the other.
Notice how unevenly the shadow shortens. The angle doubled but the shadow fell to one third, because the shadow depends on of the elevation rather than on the elevation itself. Angles and lengths do not change in proportion, which is the single most useful thing to understand about this whole chapter.
Worked example 3 — finding the angle instead. A m tall tree casts a shadow m long. Find the sun's elevation.
Worked example 4 — the difference of two shadows. A tower is m tall. Find how much longer its shadow is at an elevation of than at .
The difference is m.
The modelling assumption worth stating. These questions treat the pole as vertical and the ground as horizontal, so the triangle really is right-angled. On a slope, or with a leaning pole, the triangle is not right-angled and this method does not apply — **which is why every such question says on level ground.**
Worked example 1. A pole casts a shadow m long when the sun's elevation is . Find the height of the pole.
Worked example 2 — the same pole later in the day. Find the shadow of that pole when the sun's elevation rises to .
**So the shadow shrank from m to m** while the sun climbed from to . The height never changed, of course — and that fixed height is what lets you use one calculation to feed the other.
Notice how unevenly the shadow shortens. The angle doubled but the shadow fell to one third, because the shadow depends on of the elevation rather than on the elevation itself. Angles and lengths do not change in proportion, which is the single most useful thing to understand about this whole chapter.
Worked example 3 — finding the angle instead. A m tall tree casts a shadow m long. Find the sun's elevation.
Worked example 4 — the difference of two shadows. A tower is m tall. Find how much longer its shadow is at an elevation of than at .
The difference is m.
The modelling assumption worth stating. These questions treat the pole as vertical and the ground as horizontal, so the triangle really is right-angled. On a slope, or with a leaning pole, the triangle is not right-angled and this method does not apply — **which is why every such question says on level ground.**
How do you find the right triangle hidden inside a larger figure?
Look for a perpendicular — given or constructed — and work inside one of the triangles it creates. A figure with no right angle has to be cut into pieces that have one.
Worked example 1 — a rhombus. A rhombus has side cm and one angle of . Find both diagonals.
The diagonals of a rhombus bisect each other at right angles and bisect the vertex angles, so each quarter of the figure is a right-angled triangle with hypotenuse cm and an angle of at the vertex:
So the diagonals are cm and cm.
Check by the area, two ways: cm, and base times height gives cm, as required.
Worked example 2 — two triangles sharing an altitude. In , is the perpendicular from to , with , and cm. Find .
Work in each right-angled triangle separately. In ,
and in ,
So cm.
The altitude was the bridge between the two triangles, appearing as the opposite side in both. Whenever a question gives two angles and one length across a non-right triangle, look for the altitude — it is almost always the shared side.
Worked example 3 — a trapezium. In an isosceles trapezium the parallel sides are cm and cm, and each base angle is . Find the height and the slant side.
Drop perpendiculars from the ends of the shorter side. The two horizontal offsets are equal and share the extra length:
Check the area two ways: cm, and splitting into a rectangle plus two triangles of cm gives cm, as required.
The general instruction for this section. Name the triangle you are working in before writing a ratio, and say which side is the hypotenuse in that triangle. In worked example 2 the side was opposite the angle in both triangles but the hypotenuses were different — and a ratio written for the wrong triangle is the most common way to lose a whole question here.
Worked example 1 — a rhombus. A rhombus has side cm and one angle of . Find both diagonals.
The diagonals of a rhombus bisect each other at right angles and bisect the vertex angles, so each quarter of the figure is a right-angled triangle with hypotenuse cm and an angle of at the vertex:
So the diagonals are cm and cm.
Check by the area, two ways: cm, and base times height gives cm, as required.
Worked example 2 — two triangles sharing an altitude. In , is the perpendicular from to , with , and cm. Find .
Work in each right-angled triangle separately. In ,
and in ,
So cm.
The altitude was the bridge between the two triangles, appearing as the opposite side in both. Whenever a question gives two angles and one length across a non-right triangle, look for the altitude — it is almost always the shared side.
Worked example 3 — a trapezium. In an isosceles trapezium the parallel sides are cm and cm, and each base angle is . Find the height and the slant side.
Drop perpendiculars from the ends of the shorter side. The two horizontal offsets are equal and share the extra length:
Check the area two ways: cm, and splitting into a rectangle plus two triangles of cm gives cm, as required.
The general instruction for this section. Name the triangle you are working in before writing a ratio, and say which side is the hypotenuse in that triangle. In worked example 2 the side was opposite the angle in both triangles but the hypotenuses were different — and a ratio written for the wrong triangle is the most common way to lose a whole question here.
Exam tip
What layout keeps a trigonometry word problem safe?
Draw the figure, mark the right angle, and label the known side and the known angle before writing anything else. Almost every lost mark in this chapter is a labelling error, not an arithmetic one.
- Draw the ground horizontal and the object vertical, and mark the angle at the correct vertex. An angle of elevation sits at the observer's eye, not at the top of the object
- State the equality for a depression: the angle of depression from the top equals the angle of elevation from the boat (alternate angles). It is a step worth a mark
- Choose the ratio by what you have and what you want — and write it as a named ratio first: *in , *
- Name the triangle whenever a figure contains more than one
- Keep surds exact through the working and give a decimal only at the end, using and
- Carry units into the answer, and check the answer is physically sensible: a string is longer than the height, a shadow shortens as the sun rises
- Check with the Pythagoras theorem once two sides are known, and check that the longest side faces the largest angle
The misconception to name. Doubling the angle does not double the length. In the shadow example the elevation went from to and the shadow fell from m to m — a factor of three, not two. Ratios are not proportional to angles, so every length must be recomputed rather than scaled.
- Draw the ground horizontal and the object vertical, and mark the angle at the correct vertex. An angle of elevation sits at the observer's eye, not at the top of the object
- State the equality for a depression: the angle of depression from the top equals the angle of elevation from the boat (alternate angles). It is a step worth a mark
- Choose the ratio by what you have and what you want — and write it as a named ratio first: *in , *
- Name the triangle whenever a figure contains more than one
- Keep surds exact through the working and give a decimal only at the end, using and
- Carry units into the answer, and check the answer is physically sensible: a string is longer than the height, a shadow shortens as the sun rises
- Check with the Pythagoras theorem once two sides are known, and check that the longest side faces the largest angle
The misconception to name. Doubling the angle does not double the length. In the shadow example the elevation went from to and the shadow fell from m to m — a factor of three, not two. Ratios are not proportional to angles, so every length must be recomputed rather than scaled.
Did you know
How can you measure a tree with a metre stick and a shadow?
You cannot reach the top of a tree with a tape, but its shadow lies at your feet. That is enough.
Stand a metre stick upright beside the tree and measure both shadows at the same moment. The sun's rays are effectively parallel, so both the stick and the tree make the same angle of elevation with the ground, and therefore the same tangent:
Worked example. A m stick casts a shadow of m while the tree's shadow is m long. Then
No angle had to be measured at all — the two tangents were equal, so the angle cancelled out. The sun's elevation was in fact about , but the method never needed to know it.
And there is one moment in the day when the arithmetic vanishes entirely. When the sun's elevation is exactly , , so
for every vertical object at once. Watch your own shadow, and the instant it is exactly as long as you are tall, the shadow of the tree, the pole and the water tank all equal their heights too. One glance then measures the whole neighbourhood.
The same idea, run backwards, is how a sundial works. A fixed vertical rod's shadow length is set entirely by the sun's elevation, so the length of the shadow is a reading of the time of day — and its direction is another. A shadow is a tangent drawn on the ground, and this chapter is about learning to read it.
Stand a metre stick upright beside the tree and measure both shadows at the same moment. The sun's rays are effectively parallel, so both the stick and the tree make the same angle of elevation with the ground, and therefore the same tangent:
Worked example. A m stick casts a shadow of m while the tree's shadow is m long. Then
No angle had to be measured at all — the two tangents were equal, so the angle cancelled out. The sun's elevation was in fact about , but the method never needed to know it.
And there is one moment in the day when the arithmetic vanishes entirely. When the sun's elevation is exactly , , so
for every vertical object at once. Watch your own shadow, and the instant it is exactly as long as you are tall, the shadow of the tree, the pole and the water tank all equal their heights too. One glance then measures the whole neighbourhood.
The same idea, run backwards, is how a sundial works. A fixed vertical rod's shadow length is set entirely by the sun's elevation, so the length of the shadow is a reading of the time of day — and its direction is another. A shadow is a tangent drawn on the ground, and this chapter is about learning to read it.
Exam relevance
How does solving right triangles prepare you for JEE and NEET?
This is foundation work whose successor chapter is examined in both boards and competitive papers, and whose method is used in Physics from the first week of Class 11.
Where it leads in Mathematics. Class 10 extends this to Heights and Distances, where two observation points and two angles appear in one figure, and the technique is exactly worked example 2 above — two right-angled triangles sharing an altitude. Class 11 generalises it to any triangle through the Sine Rule and Cosine Rule in Properties of Triangles, a JEE Main topic. The habit of naming the triangle before writing a ratio is what makes those chapters manageable.
Where it leads in Physics. Every inclined-plane question resolves the weight into along the slope and perpendicular to it — and identifying which is which is the same opposite or adjacent decision you make here. Projectile motion splits the launch velocity the same way, and optics uses the tangent for the apparent shift of an object in a medium. Both JEE Main and NEET set numericals whose only geometry is this one triangle.
Question types to expect. At this level: solve a triangle, find a height or a distance, work inside a figure. In competitive papers: a two-angle heights-and-distances problem, a resolved-component calculation, or an assertion-reason item on whether a given set of data determines a triangle.
That last one is worth thinking about now. Two angles do not determine a right-angled triangle's size, but one side with one angle does, and so do two sides. The same counting argument returns in Class 11 as the ambiguous case of the sine rule — where two sides and a non-included angle can give two different triangles, exactly the SSA situation from the congruence chapter.
The single trap that costs marks. Confusing elevation with depression, or placing the angle at the wrong vertex. In a resolved-components question the equivalent error is using where belongs, which flips the whole answer. Mark the angle on the figure and identify the hypotenuse of that particular triangle before choosing a ratio.
Board versus competitive emphasis. ICSE marks the figure, the named ratio and the units; a competitive paper marks one number, often as a component inside a longer physics problem. The transferable skill is the choice of ratio, and it is worth practising until it needs no thought.
Where it leads in Mathematics. Class 10 extends this to Heights and Distances, where two observation points and two angles appear in one figure, and the technique is exactly worked example 2 above — two right-angled triangles sharing an altitude. Class 11 generalises it to any triangle through the Sine Rule and Cosine Rule in Properties of Triangles, a JEE Main topic. The habit of naming the triangle before writing a ratio is what makes those chapters manageable.
Where it leads in Physics. Every inclined-plane question resolves the weight into along the slope and perpendicular to it — and identifying which is which is the same opposite or adjacent decision you make here. Projectile motion splits the launch velocity the same way, and optics uses the tangent for the apparent shift of an object in a medium. Both JEE Main and NEET set numericals whose only geometry is this one triangle.
Question types to expect. At this level: solve a triangle, find a height or a distance, work inside a figure. In competitive papers: a two-angle heights-and-distances problem, a resolved-component calculation, or an assertion-reason item on whether a given set of data determines a triangle.
That last one is worth thinking about now. Two angles do not determine a right-angled triangle's size, but one side with one angle does, and so do two sides. The same counting argument returns in Class 11 as the ambiguous case of the sine rule — where two sides and a non-included angle can give two different triangles, exactly the SSA situation from the congruence chapter.
The single trap that costs marks. Confusing elevation with depression, or placing the angle at the wrong vertex. In a resolved-components question the equivalent error is using where belongs, which flips the whole answer. Mark the angle on the figure and identify the hypotenuse of that particular triangle before choosing a ratio.
Board versus competitive emphasis. ICSE marks the figure, the named ratio and the units; a competitive paper marks one number, often as a component inside a longer physics problem. The transferable skill is the choice of ratio, and it is worth practising until it needs no thought.
Key takeaways
What should you be able to solve before moving on?
One triangle, two given parts, and a fixed procedure.
- Two parts solve a right triangle, as long as they are not both angles
- The third angle is minus the known acute angle
- Choose the ratio that contains the two knowns and the one unknown, and write it as a named ratio before substituting
- Sine for opposite and hypotenuse, cosine for adjacent and hypotenuse, tangent for the two legs
- Angle of elevation is measured up from the horizontal at the observer; angle of depression is measured down, and the two are equal for the same pair of points
- Shadow problems use , and the height stays fixed while the shadow changes
- Inside a larger figure, find or construct a perpendicular and name the triangle you are working in
- Check with the Pythagoras theorem, with the longest-side rule, and by asking whether the answer is physically sensible
- Lengths are not proportional to angles — doubling an angle does not double a side
The sharpest self-test is worked example 2 of the last section: an altitude of cm with base angles of and . Find from a blank page, and check whether you kept the two triangles separate all the way through.
- Two parts solve a right triangle, as long as they are not both angles
- The third angle is minus the known acute angle
- Choose the ratio that contains the two knowns and the one unknown, and write it as a named ratio before substituting
- Sine for opposite and hypotenuse, cosine for adjacent and hypotenuse, tangent for the two legs
- Angle of elevation is measured up from the horizontal at the observer; angle of depression is measured down, and the two are equal for the same pair of points
- Shadow problems use , and the height stays fixed while the shadow changes
- Inside a larger figure, find or construct a perpendicular and name the triangle you are working in
- Check with the Pythagoras theorem, with the longest-side rule, and by asking whether the answer is physically sensible
- Lengths are not proportional to angles — doubling an angle does not double a side
The sharpest self-test is worked example 2 of the last section: an altitude of cm with base angles of and . Find from a blank page, and check whether you kept the two triangles separate all the way through.