Opposite, adjacent and hypotenuse
In a right triangle, how do you name each side by where it sits relative to one acute angle, so that a rule about that angle can say exactly which side it means?
Where you are 7 in Right triangle trigonometry
Why this exists
A right triangle has one square corner and two sharp ones. Right-triangle trigonometry is built from statements about one of the sharp corners, each of the form “this side compared with that side”. Such a statement needs a way to point at a side that works in every right triangle, however it is drawn, turned or lettered. “The bottom side” fails as soon as the triangle is turned over, and calling it \overline{BC} fails as soon as the letters change.
The way out is to name each side by where it sits relative to the angle being looked at: across from it, beside it, or across from the right angle. The price is that two of those names belong to the angle rather than to the side: look from the other sharp corner and the same side gets a different name.
Definition
Start from the words of parts of a triangle. In a right triangle the two sides that form the right angle are the legs, and the side opposite the right angle is the hypotenuse. A triangle has at most one right angle (classifying triangles by angles), so the hypotenuse is one fixed side. The other two angles are acute: the three angles add to 180^\circ (the triangle angle sum) and the right angle takes 90^\circ, so the other two add to 90^\circ and each is less than 90^\circ.
Now choose one of the two acute angles. Let \triangle ABC have its right angle at C, and choose \angle A. Relative to \angle A, the three sides are named as follows.
- The hypotenuse is the side opposite the right angle, \overline{AB}. It is the same side whichever acute angle was chosen.
- The side opposite \angle A is the side that does not have A as an endpoint, \overline{BC}.
- The side adjacent to \angle A is the leg that has A as an endpoint, \overline{AC}. Put another way, it is the one side of \angle A that is not the hypotenuse.
“Adjacent” means “next to”. Two sides meet at A, \overline{AC} and \overline{AB}, and in everyday speech both are next to the angle. But \overline{AB} already has its own name, the hypotenuse, so “adjacent” is kept for the leg. (In adjacent angles the word describes two angles that share a side, a different use.)
The lengths of the three sides are often shortened to \text{opp}, \text{adj} and \text{hyp}, and an acute angle with no letter of its own is often called \theta. These names mean something only once a right triangle and one of its acute angles are fixed.
Intuition
Stand at the corner A and look into the triangle. Two sides start at your feet: they are the two arms of your angle. One of them runs to the square corner; that is the adjacent side. The other runs to the far sharp corner; that is the hypotenuse. The third side does not reach you at all. It lies across the triangle, and it is the opposite side.
Now walk to B, as in Figure 2.
The hypotenuse still starts at your feet, because it joins the two sharp corners and you are standing at one of them. The other two sides trade places. \overline{BC}, which lay across the triangle when you stood at A, now starts at your feet and runs to the square corner, so it is adjacent. \overline{AC}, which started at your feet, is now the side you cannot touch, so it is opposite.
Nothing about the triangle changed between Figure 1 and Figure 2. Only the angle changed, and two of the three names changed with it. “Opposite” and “adjacent” are not properties of a side. Each one is a relationship between a side and an angle.
Results
In each fact below the triangle is a right triangle and the angle is one of its two acute angles.
Fact 1 (one name each). Each of the three sides is exactly one of these: the hypotenuse, the side opposite the angle, the side adjacent to the angle.
Fact 2 (the hypotenuse is shared). Both acute angles have the same hypotenuse.
Fact 3 (the legs swap). If \alpha and \beta are the two acute angles, the side opposite \alpha is the side adjacent to \beta, and the side adjacent to \alpha is the side opposite \beta.
For \triangle ABC with its right angle at C:
| Side | named from \angle A | named from \angle B |
|---|---|---|
| \overline{AB} | hypotenuse | hypotenuse |
| \overline{BC} | opposite | adjacent |
| \overline{AC} | adjacent | opposite |
Fact 4 (the Pythagorean theorem in these names). For either acute angle,
Fact 5 (the hypotenuse is longest). The hypotenuse is longer than the opposite side and longer than the adjacent side.
Fact 6 (which leg is longer). For an acute angle \theta of a right triangle:
| If \theta is | then |
|---|---|
| less than 45^\circ | \text{opp} < \text{adj} |
| exactly 45^\circ | \text{opp} = \text{adj} |
| more than 45^\circ | \text{opp} > \text{adj} |
Each row works in reverse too: the comparison of the two legs tells you which row \theta is in.
Why it is true
Throughout, \triangle ABC has its right angle at C. Since A and B play the same part, it is enough to argue from \angle A.
Fact 1. The sides are \overline{AB}, \overline{BC} and \overline{CA}. Exactly one of them, \overline{BC}, does not have A as an endpoint, so it is the opposite side and neither of the others is. The remaining two both end at A. Of these, \overline{AB} does not have C as an endpoint, so it is opposite the right angle: the hypotenuse. That leaves \overline{AC}, which ends at A and at the right angle, so it is a leg with A as an endpoint: the adjacent side. Each side received one name and no side received two.
Fact 2. The hypotenuse is defined as the side opposite the right angle. The acute angle that was chosen plays no part in that definition, so both acute angles have \overline{AB} as their hypotenuse.
Fact 3. \overline{BC} does not end at A, so it is opposite \angle A. It does end at B, and it ends at C too, so it is a leg; a leg ending at B is the side adjacent to \angle B. In the same way, \overline{AC} does not end at B, so it is opposite \angle B, and it is a leg ending at A, so it is adjacent to \angle A.
Fact 4. By Fact 1, the two sides other than the hypotenuse are the opposite and the adjacent side, and those two are the legs. The Pythagorean theorem says the squares of the legs add to the square of the hypotenuse.
Fact 5. Lengths are positive, so \text{adj}^2 > 0 and Fact 4 gives \text{hyp}^2 > \text{opp}^2. For positive numbers the larger square belongs to the larger number, so \text{hyp} > \text{opp}. The same argument with the two legs exchanged gives \text{hyp} > \text{adj}.
Fact 6. Call the other acute angle \varphi. The two acute angles add to 90^\circ, so {\varphi = 90^\circ - \theta}. Then {\theta > \varphi} exactly when {\theta > 90^\circ - \theta}, that is, when {2\theta > 90^\circ}, which is when {\theta > 45^\circ}. Likewise {\theta = \varphi} exactly when {\theta = 45^\circ}.
In any triangle the longer of two sides lies opposite the larger of the two angles, and the reverse also holds (sides and angles in order). Equal sides lie opposite equal angles and the reverse also holds (the base angles theorem and its converse). The side opposite \theta is \text{opp}, and by Fact 3 the side opposite \varphi is \text{adj}. So \text{opp} > \text{adj} exactly when \theta > \varphi, which is exactly when \theta > 45^\circ. The equal case and the smaller case go the same way.
Worked examples
A triangle resting on its hypotenuse
In Figure 3, \triangle PQR has its right angle at Q. Name the hypotenuse, then the sides opposite and adjacent to \angle P, then the sides opposite and adjacent to \angle R.
The hypotenuse is the side opposite the right angle, the one side without Q as an endpoint: \overline{PR}. It lies along the bottom of the picture, which is not where the eye expects it, but the definition only asks which side is across from the right angle.
For \angle P, the opposite side is the side without P: \overline{QR}. The adjacent side is the leg that ends at P: \overline{PQ}.
For \angle R, the opposite side is the side without R: \overline{PQ}. The adjacent side is the leg that ends at R: \overline{QR}.
Check. As Fact 3 says, \overline{QR} is opposite \angle P and adjacent to \angle R, and \overline{PQ} is the other way round.
The same triangle from both angles
In \triangle LMN the right angle is at N, LN = 20 and MN = 21. Find LM. Then give the lengths of the opposite side, the adjacent side and the hypotenuse, first for \angle L and then for \angle M.
Solution\overline{LM} is the side without N, so it is the hypotenuse, and the other two sides are the legs. By the Pythagorean theorem,
so LM = 29, the positive square root, since a length is positive.
For \angle L, the opposite side is the side without L, which is \overline{MN}, of length 21. The adjacent side is the leg ending at L, which is \overline{LN}, of length 20.
For \angle M, the opposite side is \overline{LN}, of length 20, and the adjacent side is \overline{MN}, of length 21.
| Side | from \angle L | from \angle M |
|---|---|---|
| opposite | 21 | 20 |
| adjacent | 20 | 21 |
| hypotenuse | 29 | 29 |
Fact 6 adds one more thing for free. From \angle L the opposite side is the longer leg, so \angle L is more than 45^\circ, and \angle M is less than 45^\circ.
A missing side with a square root
For an acute angle \theta of a right triangle, the hypotenuse is 9 and the side opposite \theta is 4. Find the side adjacent to \theta. Is \theta more or less than 45^\circ?
SolutionThe opposite and adjacent sides are the two legs, so Fact 4 applies:
The negative root is discarded because a length is positive.
To compare the legs, compare their squares: \text{opp}^2 = 16 and \text{adj}^2 = 65. Since 16 < 65, the opposite side is the shorter leg, and by Fact 6, \theta is less than 45^\circ. No decimal was needed.
Seen from the other acute angle, the same two legs swap: its opposite side is \displaystyle \sqrt{65} and its adjacent side is 4.
Sides written with a variable
For an acute angle \theta of a right triangle, the adjacent side is x, the opposite side is x + 7 and the hypotenuse is x + 8. Find x and the three lengths, and decide whether \theta is more or less than 45^\circ.
SolutionThe two legs are the opposite and adjacent sides, so by Fact 4,
So x = 5 or x = -3. The adjacent side has length x, and a length must be positive, so the root -3 is thrown out and x = 5.
The adjacent side is 5, the opposite side is {5 + 7 = 12} and the hypotenuse is {5 + 8 = 13}.
Check. {5^2 + 12^2 = 25 + 144 = 169 = 13^2}, and the hypotenuse is the longest of the three, as Fact 5 requires.
The opposite side is the longer leg, so \theta is more than 45^\circ.
One side, three different names
In Figure 4, \triangle ABC has its right angle at C, and \overline{CD} is perpendicular to \overline{AB}, with D on \overline{AB}.
(a) In \triangle ABC, name the hypotenuse and the sides opposite and adjacent to \angle A. (b) Do the same for \angle A in \triangle ACD. (c) Name the hypotenuse of \triangle BCD and the sides opposite and adjacent to \angle B in it. (d) Which segments change their name from one triangle to another?
Solution(a) The right angle is at C, so the hypotenuse is \overline{AB}. The side without A is \overline{BC}, the opposite side. The leg ending at A is \overline{AC}, the adjacent side.
(b) \triangle ACD has its right angle at D, because \overline{CD} is perpendicular to \overline{AB}. Its angle at A is the same angle as before, since D lies on \overline{AB} and so the ray from A through D is the ray from A through B. The hypotenuse is the side without D: \overline{AC}. The side opposite \angle A is the side without A: \overline{CD}. The side adjacent to \angle A is the leg ending at A: \overline{AD}.
(c) \triangle BCD also has its right angle at D. Its hypotenuse is the side without D: \overline{BC}. The side opposite \angle B is \overline{CD}, and the side adjacent to \angle B is \overline{BD}.
(d) \overline{AC} carries all three names. In \triangle ABC it is adjacent to \angle A and, by Fact 3, opposite \angle B; in \triangle ACD it is the hypotenuse. \overline{BC} does the same: opposite \angle A and adjacent to \angle B in \triangle ABC, the hypotenuse of \triangle BCD. \overline{CD} is not a side of \triangle ABC at all, but it is opposite \angle A in one small triangle and opposite \angle B in the other.
The same angle, \angle A, has a different opposite side, a different adjacent side and a different hypotenuse in the two triangles it belongs to. A side’s name depends on the angle and on the triangle, so both have to be said.
On a coordinate grid
The points {O = (0, 0)}, {F = (12, 0)} and {P = (12, 5)} make a triangle. (a) Show that it is a right triangle and say where the right angle is. (b) Give the lengths of the opposite side, the adjacent side and the hypotenuse for the angle at O. (c) Do the same for the angle at P.
Solution(a) \overline{OF} lies along the x-axis, so it is horizontal. F and P both have x-coordinate 12, so \overline{FP} is vertical. A horizontal and a vertical segment are perpendicular, so the right angle is at F.
(b) The hypotenuse is the side without F, which is \overline{OP}. The legs have lengths OF = 12 and FP = 5, so
The side opposite the angle at O is the side without O, which is \overline{FP}, of length 5. The side adjacent to it is the leg ending at O, which is \overline{OF}, of length 12.
(c) For the angle at P, the opposite side is \overline{OF}, of length 12, the adjacent side is \overline{FP}, of length 5, and the hypotenuse is still 13.
The angle at O is the angle \overline{OP} makes with the x-axis. Its adjacent leg is horizontal, so it measures the run from O to P (the change in x), and its opposite leg is vertical, so it measures the rise (the change in y). Here P is up and to the right of O, so the slope of \overline{OP}, rise over run, equals the opposite side over the adjacent side, \tfrac{5}{12}. Two conditions made that work. The angle had to be the one at the horizontal leg: from the angle at P the names swap, and opposite over adjacent is \tfrac{12}{5}, run over rise. And the segment had to go uphill: had it run from (0, 5) down to (12, 0), the side lengths would be the same positive numbers but the slope would be -\tfrac{5}{12}. Comparing the opposite side with the adjacent side for any acute angle is the subject of the tangent ratio.
Boundaries
Right triangles only. A hypotenuse exists only when there is a right angle. In a triangle with no right angle each angle still has an opposite side, as in parts of a triangle, but both of the other sides meet at the angle and nothing sets one of them apart, so “the adjacent side” names no single side.
Acute angles only. At the right angle itself, the opposite side is the hypotenuse, and the two sides that meet there are both legs. Both are next to it and neither is singled out, so “the side adjacent to the right angle” is ambiguous and is left undefined. “Opposite” and “adjacent” in this lesson’s sense are defined only for the two acute angles.
A name needs its angle and its triangle. “The opposite side” with no angle attached means nothing, and Example 5 shows that even with the angle fixed, a segment can be a leg of one triangle and the hypotenuse of another.
The right angle must be known. A corner that looks square may measure 89^\circ, and then there is no hypotenuse. A right angle counts only when it is marked with a small square or stated (what a diagram may assume).
Where this goes next
The three names exist so that the sides of a right triangle can be compared in a way that depends only on the angle. The sine ratio compares the opposite side with the hypotenuse, the cosine ratio compares the adjacent side with the hypotenuse, and the tangent ratio compares the opposite side with the adjacent side. The three ratios turned upside down are secant, cosecant and cotangent. Why each ratio depends only on the size of the angle, and not on how big the triangle is, is the subject of why a trig ratio depends only on the angle.
Fact 3, that the legs swap between the two acute angles, is the whole reason behind why sine and cosine swap in a right triangle.
Every problem in finding a missing side with a trig ratio and solving right triangles starts with the step this lesson teaches: stand at the known angle and name the sides from there. The special right triangles are the ones where those names come with exact lengths.
Exercises
Naming from the letters
- In \triangle RST the right angle is at S. Name the hypotenuse.
- In \triangle RST, with the right angle at S, name the side opposite \angle R and the side adjacent to \angle R.
- In the same triangle, name the side opposite \angle T and the side adjacent to \angle T.
- In \triangle GHK the right angle is at K. The side \overline{GK} is adjacent to one acute angle and opposite the other. Which is which?
- In \triangle UVW the right angle is at V. Which acute angle has \overline{UV} as its opposite side?
Two right triangles in one figure
Use Figure 5 for Exercises 6 to 9. The right angle is at C and D lies on \overline{CA}.
- Name the two right triangles in Figure 5 and say where the right angle is in each.
- In \triangle DBC, name the hypotenuse, the side opposite \angle BDC and the side adjacent to \angle BDC.
- In \triangle ABC, which acute angle is \overline{BC} opposite, and which is it adjacent to? Answer the same question for \triangle DBC.
- The third triangle in the figure, \triangle ABD, has no hypotenuse. Explain why.
Finding lengths
- For an acute angle \theta of a right triangle, the opposite side is 7 and the adjacent side is 24. Find the hypotenuse.
- For an acute angle \theta of a right triangle, the hypotenuse is 61 and the adjacent side is 11. Find the opposite side.
- For an acute angle \theta of a right triangle, the hypotenuse is 6 and the opposite side is 2. Find the adjacent side exactly.
- For an acute angle \theta of a right triangle, the opposite and adjacent sides are both 3. Find the hypotenuse, and find \theta.
- In \triangle ABC the right angle is at C, AB = 26 and BC = 10. Find AC, then give the lengths of the opposite side, the adjacent side and the hypotenuse for \angle A and for \angle B.
- For an acute angle \theta of a right triangle, the opposite side is x, the adjacent side is x + 31 and the hypotenuse is x + 32. Find x and the three lengths.
- The points {O = (0, 0)}, {F = (35, 0)} and {P = (35, 12)} make a right triangle. Give the lengths of the opposite side, the adjacent side and the hypotenuse for the angle at O.
Which leg is longer
- One acute angle of a right triangle measures 38^\circ. Which is longer, the side opposite it or the side adjacent to it?
- For an acute angle \theta of a right triangle, the opposite side is 11 and the hypotenuse is 14. Find the adjacent side exactly, and decide whether \theta is more or less than 45^\circ.
Reasoning
- Explain why “the side opposite the right angle” names a single side of a right triangle but “the side adjacent to the right angle” does not.
- Explain why the side opposite one acute angle of a right triangle is always the side adjacent to the other acute angle.
- Explain why the two acute angles of a right triangle cannot both have an opposite side longer than their adjacent side.
- Prove that for an acute angle \theta of a right triangle, the opposite side and the adjacent side are equal exactly when \theta = 45^\circ.
Answers
- \overline{RT}. The hypotenuse is opposite the right angle, so it is the side without S.
- Opposite \angle R: \overline{ST}, the side without R. Adjacent to \angle R: \overline{RS}, the leg that ends at R.
- Opposite \angle T: \overline{RS}. Adjacent to \angle T: \overline{ST}. The two legs have swapped names, as Fact 3 says.
- \overline{GK} is adjacent to \angle G and opposite \angle H. It ends at G and at the right angle K, so it is a leg ending at G; it does not end at H.
- \angle W. \overline{UV} does not have W as an endpoint. (It is adjacent to \angle U.)
- \triangle ABC and \triangle DBC, both with the right angle at C. D lies on \overline{CA}, so the angle \angle BCD is the angle \angle BCA, which is a right angle.
- Hypotenuse: \overline{BD}, the side without C. Opposite \angle BDC: \overline{BC}, the side without D. Adjacent to \angle BDC: \overline{CD}, the leg ending at D.
- In \triangle ABC, \overline{BC} is opposite \angle A and adjacent to \angle ABC. In \triangle DBC, it is opposite \angle BDC and adjacent to \angle DBC. In each triangle it is a leg ending at B and missing the other acute vertex.
- \triangle ABD has no right angle. \angle A is an acute angle of right \triangle ABC. \angle ABD is part of \angle ABC, which is acute, so it is acute too. \angle ADB forms a linear pair with \angle BDC, an acute angle of right \triangle DBC, so \angle ADB is more than 90^\circ. With no right angle there is no hypotenuse.
- 25. By Fact 4, \text{hyp}^2 = 7^2 + 24^2 = 49 + 576 = 625, and \displaystyle \sqrt{625} = 25.
- 60. \text{opp}^2 = 61^2 - 11^2 = 3721 - 121 = 3600, and \displaystyle \sqrt{3600} = 60.
- \displaystyle 4\sqrt{2}. \text{adj}^2 = 6^2 - 2^2 = 36 - 4 = 32, and \displaystyle \sqrt{32} = \sqrt{16 \cdot 2} = 4\sqrt{2}.
- The hypotenuse is \displaystyle 3\sqrt{2} and \theta = 45^\circ. {\text{hyp}^2 = 9 + 9 = 18} and \displaystyle {\sqrt{18} = 3\sqrt{2}}. The legs are equal, so by Fact 6, \theta = 45^\circ.
- AC = 24. For \angle A: opposite 10, adjacent 24, hypotenuse 26. For \angle B: opposite 24, adjacent 10, hypotenuse 26. AC^2 = 26^2 - 10^2 = 676 - 100 = 576. The side opposite \angle A is \overline{BC} and the side adjacent to it is \overline{AC}; from \angle B they swap.
- x = 9, and the sides are 9 (opposite), 40 (adjacent) and 41 (hypotenuse). From x^2 + (x + 31)^2 = (x + 32)^2, expanding and taking x^2 from both sides gives {x^2 + 62x + 961 = 64x + 1024}, so {x^2 - 2x - 63 = 0} and {(x - 9)(x + 7) = 0}. The root -7 would make the opposite side negative. Check: {81 + 1600 = 1681 = 41^2}.
- Opposite 12, adjacent 35, hypotenuse 37. The right angle is at F, since \overline{OF} is horizontal and \overline{FP} vertical. The side without O is \overline{FP}, of length 12; the leg ending at O is \overline{OF}, of length 35; and \displaystyle OP = \sqrt{35^2 + 12^2} = \sqrt{1369} = 37.
- The adjacent side. The other acute angle is 90^\circ - 38^\circ = 52^\circ, the larger of the two, and the side opposite it is the side adjacent to the 38^\circ angle. The longer side lies opposite the larger angle. (Fact 6: 38^\circ < 45^\circ, so \text{opp} < \text{adj}.)
- \displaystyle \text{adj} = 5\sqrt{3}, and \theta is more than 45^\circ. {\text{adj}^2 = 14^2 - 11^2 = 196 - 121 = 75} and \displaystyle {\sqrt{75} = 5\sqrt{3}}. Since 75 < 121, the adjacent side is shorter than the opposite side, so by Fact 6, \theta > 45^\circ.
- A triangle has exactly one side without a given vertex, so the side opposite the right angle is one side, the hypotenuse. Two sides meet at the right angle, and both are legs, so there is nothing to pick one of them out. At an acute angle the two sides that meet there are one leg and the hypotenuse, and “adjacent” names the leg.
- Put the right angle at C and the acute angles at A and B. The side opposite \angle A is the side without A, which is \overline{BC}. It ends at C, so it is a leg, and it ends at B, so it is the leg ending at B: the side adjacent to \angle B.
- Call the legs a and b and the acute angles \alpha and \beta, with a opposite \alpha. By Fact 3, a is adjacent to \beta and b is opposite \beta. If the opposite side of \alpha is longer than its adjacent side, then a > b; but then for \beta the opposite side b is shorter than the adjacent side a. The two comparisons are the same comparison of the same two legs, read in opposite directions, so at most one angle can have the longer leg opposite it.
- Let \varphi be the other acute angle, so \theta + \varphi = 90^\circ. By Fact 3 the side adjacent to \theta is the side opposite \varphi. If \theta = 45^\circ, then \varphi = 45^\circ, the angles are equal, and the sides opposite them are equal (converse of the base angles theorem), so \text{opp} = \text{adj}. If \text{opp} = \text{adj}, the angles opposite those sides, \theta and \varphi, are equal (base angles theorem), and since they add to 90^\circ, each is 45^\circ.