In a right triangle, what is the side opposite the right angle?

We can find an unknown side in a right-angled triangle when we know:

one length, and
one angle (apart from the right angle).

Example: Depth to the Seabed

The ship is anchored on the seabed.

We know:

the cable length (30 m), and
the angle the cable makes with the seabed

So we should be able to find the depth!

But How?

The answer is to use Sine, Cosine or Tangent!

But Which One?

Which one of Sine, Cosine or Tangent to use?

To find out which, first we give names to the sides:

Adjacent is adjacent (next to) to the angle,
Opposite is opposite the angle,
and the longest side is the Hypotenuse.

Now, for the side we already know and the side we are trying to find, we use the first letters of their names and the phrase "SOHCAHTOA" to decide which function:

SOH...		Sine: sin(θ) = Opposite / Hypotenuse
...CAH...		Cosine: cos(θ) = Adjacent / Hypotenuse
...TOA		Tangent: tan(θ) = Opposite / Adjacent

Like this:

Example: Depth to the Seabed (Continued)

Find the names of the two sides we are working on:

the side we know is the Hypotenuse
the side we want to find is Opposite the angle (check for yourself that "d" is opposite the angle 39°)

Now use the first letters of those two sides (Opposite and Hypotenuse) and the phrase "SOHCAHTOA" which gives us "SOHcahtoa", which tells us we need to use Sine:

Sine: sin(θ) = Opposite / Hypotenuse

Now put in the values we know:

sin(39°) = d / 30

And solve that equation!

But how do we calculate sin(39°) ... ?

Use your calculator. Type in 39 and then use the "sin" key.

That's easy!

sin(39°) = 0.6293...

So now we have:

0.6293... = d / 30

Now we rearrange it a little bit, and solve:

Start with:0.6293... = d / 30

Swap sides:d / 30 = 0.6293...

Multiply both sides by 30:d = 0.6293... x 30

Calculate:d = 18.88 to 2 decimal places

The depth the anchor ring lies beneath the hole is 18.88 m

Step By Step

These are the four steps to follow:

Step 1 Find the names of the two sides we are using, one we are trying to find and one we already know, out of Opposite, Adjacent and Hypotenuse.

Step 2 Use SOHCAHTOA to decide which one of Sine, Cosine or Tangent to use in this question.

Step 3 For Sine write down Opposite/Hypotenuse, for Cosine write down Adjacent/Hypotenuse or for Tangent write down Opposite/Adjacent. One of the values is the unknown length.

Step 4 Solve using your calculator and your skills with Algebra.

Examples

Let’s look at a few more examples:

Example: find the height of the plane.

We know the distance to the plane is 1000
And the angle is 60°

What is the plane's height?

Careful! The 60° angle is at the top, so the "h" side is Adjacent to the angle!

Step 1 The two sides we are using are Adjacent (h) and Hypotenuse (1000).

Step 2 SOHCAHTOA tells us to use Cosine.

Step 3 Put our values into the Cosine equation:
cos 60° = Adjacent / Hypotenuse
= h / 1000

Start with:cos 60° = h/1000

Swap:h/1000 = cos 60°

Calculate cos 60°:h/1000 = 0.5

Multiply both sides by 1000:h = 0.5 x 1000

h = 500

The height of the plane = 500 meters

Example: Find the length of the side y:

Step 1 The two sides we are using are Opposite (y)
and Adjacent (7).

Step 2 SOHCAHTOA tells us to use Tangent.

Step 3 Put our values into the tangent function:
tan 53° = Opposite/Adjacent
= y/7

Start with:tan 53° = y/7

Swap:y/7 = tan 53°

Multiply both sides by 7:y = 7 tan 53°

Calculate:y = 7 x 1.32704...

y = 9.29 to 2 decimal places

Side y = 9.29

Example: Radio Mast

There is a mast that is 70 meters high.

A wire goes to the top of the mast at an angle of 68°.

How long is the wire?

Step 1 The two sides we are using are Opposite (70) and Hypotenuse (w).

Step 2 SOHCAHTOA tells us to use Sine.

Step 3 Write down:
sin 68° = 70/w

The unknown length is on the bottom (the denominator) of the fraction!

So we need to follow a slightly different approach when solving:

Start with:sin 68° = 70/w

Multiply both sides by w:w × (sin 68°) = 70

Divide both sides by "sin 68°":w = 70 / (sin 68°)

Calculate:w = 70 / 0.9271...

w = 75.5 m to 1 decimal place

The length of the wire = 75.5 m

258, 1504, 1505, 1506, 1507, 2346, 2347, 2348, 3935, 248

When one angle is a 90-degree angle

A right triangle (American English) or right-angled triangle (British), or more formally an orthogonal triangle, formerly called a rectangled triangle[1] (Ancient Greek: ὀρθόσγωνία, lit. 'upright angle'),[2] is a triangle in which one angle is a right angle (that is, a 90-degree angle), i.e., in which two sides are perpendicular. The relation between the sides and other angles of the right triangle is the basis for trigonometry.

The side opposite to the right angle is called the hypotenuse (side c in the figure). The sides adjacent to the right angle are called legs (or catheti, singular: cathetus). Side a may be identified as the side adjacent to angle B and opposed to (or opposite) angle A, while side b is the side adjacent to angle A and opposed to angle B.

If the lengths of all three sides of a right triangle are integers, the triangle is said to be a Pythagorean triangle and its side lengths are collectively known as a Pythagorean triple.

Principal properties

Area

As with any triangle, the area is equal to one half the base multiplied by the corresponding height. In a right triangle, if one leg is taken as the base then the other is height, so the area of a right triangle is one half the product of the two legs. As a formula the area T is

T = 1 2 a b {\displaystyle T={\tfrac {1}{2}}ab}

where a and b are the legs of the triangle.

If the incircle is tangent to the hypotenuse AB at point P, then denoting the semi-perimeter (a + b + c) / 2 as s, we have PA = s − a and PB = s − b, and the area is given by

T = PA ⋅ PB = ( s − a ) ( s − b ) . {\displaystyle T={\text{PA}}\cdot {\text{PB}}=(s-a)(s-b).}

This formula only applies to right triangles.[3]

Altitudes

Altitude of a right triangle

If an altitude is drawn from the vertex with the right angle to the hypotenuse then the triangle is divided into two smaller triangles which are both similar to the original and therefore similar to each other. From this:

The altitude to the hypotenuse is the geometric mean (mean proportional) of the two segments of the hypotenuse.[4]: 243
Each leg of the triangle is the mean proportional of the hypotenuse and the segment of the hypotenuse that is adjacent to the leg.

In equations,

f 2 = d e , {\displaystyle \displaystyle f^{2}=de,}

(this is sometimes known as the right triangle altitude theorem)

b 2 = c e , {\displaystyle \displaystyle b^{2}=ce,}

a 2 = c d {\displaystyle \displaystyle a^{2}=cd}

where a, b, c, d, e, f are as shown in the diagram.[5] Thus

f = a b c . {\displaystyle f={\frac {ab}{c}}.}

Moreover, the altitude to the hypotenuse is related to the legs of the right triangle by[6][7]

1 a 2 + 1 b 2 = 1 f 2 . {\displaystyle {\frac {1}{a^{2}}}+{\frac {1}{b^{2}}}={\frac {1}{f^{2}}}.}

For solutions of this equation in integer values of a, b, f, and c, see here.

The altitude from either leg coincides with the other leg. Since these intersect at the right-angled vertex, the right triangle's orthocenter—the intersection of its three altitudes—coincides with the right-angled vertex.

Pythagorean theorem

The diagram for Euclid's proof of the Pythagorean theorem: each smaller square has area equal to the rectangle of corresponding color.

The Pythagorean theorem states that:

In any right triangle, the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares whose sides are the two legs (the two sides that meet at a right angle).

This can be stated in equation form as

a 2 + b 2 = c 2 {\displaystyle \displaystyle a^{2}+b^{2}=c^{2}}

where c is the length of the hypotenuse, and a and b are the lengths of the remaining two sides.

Pythagorean triples are integer values of a, b, c satisfying this equation

Inradius and circumradius

The radius of the incircle of a right triangle with legs a and b and hypotenuse c is

r = a + b − c 2 = a b a + b + c . {\displaystyle r={\frac {a+b-c}{2}}={\frac {ab}{a+b+c}}.}

The radius of the circumcircle is half the length of the hypotenuse,

R = c 2 . {\displaystyle R={\frac {c}{2}}.}

Thus the sum of the circumradius and the inradius is half the sum of the legs:[8]

R + r = a + b 2 . {\displaystyle R+r={\frac {a+b}{2}}.}

One of the legs can be expressed in terms of the inradius and the other leg as

a = 2 r ( b − r ) b − 2 r . {\displaystyle \displaystyle a={\frac {2r(b-r)}{b-2r}}.}

Characterizations

A triangle ABC with sides $a ≤ b < c {\displaystyle a\leq b<c}$

, semiperimeter s, area T, altitude h opposite the longest side, circumradius R, inradius r, exradii ra, rb, rc (tangent to a, b, c respectively), and medians ma, mb, mc is a right triangle if and only if any one of the statements in the following six categories is true. All of them are of course also properties of a right triangle, since characterizations are equivalences.

Sides and semiperimeter

$a 2 + b 2 = c 2 ( Pythagorean theorem ) {\displaystyle \displaystyle a^{2}+b^{2}=c^{2}\quad ({\text{Pythagorean theorem}})}$
$( s − a ) ( s − b ) = s ( s − c ) {\displaystyle \displaystyle (s-a)(s-b)=s(s-c)}$
$s = 2 R + r . {\displaystyle \displaystyle s=2R+r.}$ [9]
$a 2 + b 2 + c 2 = 8 R 2 . {\displaystyle \displaystyle a^{2}+b^{2}+c^{2}=8R^{2}.}$ [10]

Angles

A and B are complementary.[11]
$cos ⁡ A cos ⁡ B cos ⁡ C = 0. {\displaystyle \displaystyle \cos {A}\cos {B}\cos {C}=0.}$ [10][12]
$sin 2 ⁡ A + sin 2 ⁡ B + sin 2 ⁡ C = 2. {\displaystyle \displaystyle \sin ^{2}{A}+\sin ^{2}{B}+\sin ^{2}{C}=2.}$ [10][12]
$cos 2 ⁡ A + cos 2 ⁡ B + cos 2 ⁡ C = 1. {\displaystyle \displaystyle \cos ^{2}{A}+\cos ^{2}{B}+\cos ^{2}{C}=1.}$ [12]
$sin ⁡ 2 A = sin ⁡ 2 B = 2 sin ⁡ A sin ⁡ B . {\displaystyle \displaystyle \sin {2A}=\sin {2B}=2\sin {A}\sin {B}.}$

Area

$T = a b 2 {\displaystyle \displaystyle T={\frac {ab}{2}}}$
$T = r a r b = r r c {\displaystyle \displaystyle T=r_{a}r_{b}=rr_{c}}$
$T = r ( 2 R + r ) {\displaystyle \displaystyle T=r(2R+r)}$
$T = ( 2 s − c ) 2 − c 2 4 = s ( s − c ) {\displaystyle \displaystyle T={\frac {(2s-c)^{2}-c^{2}}{4}}=s(s-c)}$
$T = P A ⋅ P B , {\displaystyle T=PA\cdot PB,}$ where P is the tangency point of the incircle at the longest side AB.[13]

Inradius and exradii

$r = s − c = ( a + b − c ) / 2 {\displaystyle \displaystyle r=s-c=(a+b-c)/2}$
$r a = s − b = ( a − b + c ) / 2 {\displaystyle \displaystyle r_{a}=s-b=(a-b+c)/2}$
$r b = s − a = ( − a + b + c ) / 2 {\displaystyle \displaystyle r_{b}=s-a=(-a+b+c)/2}$
$r c = s = ( a + b + c ) / 2 {\displaystyle \displaystyle r_{c}=s=(a+b+c)/2}$
$r a + r b + r c + r = a + b + c {\displaystyle \displaystyle r_{a}+r_{b}+r_{c}+r=a+b+c}$
$r a 2 + r b 2 + r c 2 + r 2 = a 2 + b 2 + c 2 {\displaystyle \displaystyle r_{a}^{2}+r_{b}^{2}+r_{c}^{2}+r^{2}=a^{2}+b^{2}+c^{2}}$
$r = r a r b r c . {\displaystyle \displaystyle r={\frac {r_{a}r_{b}}{r_{c}}}.}$ [14]

Altitude and medians

The altitude of a right triangle from its right angle to its hypotenuse is the geometric mean of the lengths of the segments the hypotenuse is split into. Using Pythagoras' theorem on the 3 triangles of sides (p + q, r, s ), (r, p, h ) and (s, h, q ),

( p + q ) 2 = r 2 + s 2 p 2 + 2 p q + q 2 = p 2 + h 2 ⏞ + h 2 + q 2 ⏞ 2 p q = 2 h 2 ∴ h = p q {\displaystyle {\begin{aligned}(p+q)^{2}\;\;&=\quad r^{2}\;\;\,+\quad s^{2}\\p^{2}\!\!+\!2pq\!+\!q^{2}&=\overbrace {p^{2}\!\!+\!h^{2}} +\overbrace {h^{2}\!\!+\!q^{2}} \\2pq\quad \;\;\;&=2h^{2}\;\therefore h\!=\!{\sqrt {pq}}\\\end{aligned}}}

$h = a b c {\displaystyle \displaystyle h={\frac {ab}{c}}}$
$m a 2 + m b 2 + m c 2 = 6 R 2 . {\displaystyle \displaystyle m_{a}^{2}+m_{b}^{2}+m_{c}^{2}=6R^{2}.}$ [8]: Prob. 954, p. 26
The length of one median is equal to the circumradius.
The shortest altitude (the one from the vertex with the biggest angle) is the geometric mean of the line segments it divides the opposite (longest) side into. This is the right triangle altitude theorem.

Circumcircle and incircle

The triangle can be inscribed in a semicircle, with one side coinciding with the entirety of the diameter (Thales' theorem).
The circumcenter is the midpoint of the longest side.
The longest side is a diameter of the circumcircle $( c = 2 R ) . {\displaystyle \displaystyle (c=2R).}$
The circumcircle is tangent to the nine-point circle.[10]
The orthocenter lies on the circumcircle.[8]
The distance between the incenter and the orthocenter is equal to $2 r {\displaystyle {\sqrt {2}}r}$ .[8]

Trigonometric ratios

The trigonometric functions for acute angles can be defined as ratios of the sides of a right triangle. For a given angle, a right triangle may be constructed with this angle, and the sides labeled opposite, adjacent and hypotenuse with reference to this angle according to the definitions above. These ratios of the sides do not depend on the particular right triangle chosen, but only on the given angle, since all triangles constructed this way are similar. If, for a given angle α, the opposite side, adjacent side and hypotenuse are labeled O, A and H respectively, then the trigonometric functions are

sin ⁡ α = O H , cos ⁡ α = A H , tan ⁡ α = O A , sec ⁡ α = H A , cot ⁡ α = A O , csc ⁡ α = H O . {\displaystyle \sin \alpha ={\frac {O}{H}},\,\cos \alpha ={\frac {A}{H}},\,\tan \alpha ={\frac {O}{A}},\,\sec \alpha ={\frac {H}{A}},\,\cot \alpha ={\frac {A}{O}},\,\csc \alpha ={\frac {H}{O}}.}

For the expression of hyperbolic functions as ratio of the sides of a right triangle, see the hyperbolic triangle of a hyperbolic sector.

Special right triangles

The values of the trigonometric functions can be evaluated exactly for certain angles using right triangles with special angles. These include the 30-60-90 triangle which can be used to evaluate the trigonometric functions for any multiple of π/6, and the 45-45-90 triangle which can be used to evaluate the trigonometric functions for any multiple of π/4.

Kepler triangle

Let H, G, and A be the harmonic mean, the geometric mean, and the arithmetic mean of two positive numbers a and b with a > b. If a right triangle has legs H and G and hypotenuse A, then[15]

A H = A 2 G 2 = G 2 H 2 = ϕ {\displaystyle {\frac {A}{H}}={\frac {A^{2}}{G^{2}}}={\frac {G^{2}}{H^{2}}}=\phi \,}

and

a b = ϕ 3 , {\displaystyle {\frac {a}{b}}=\phi ^{3},\,}

where $ϕ {\displaystyle \phi }$

is the golden ratio

1 + 5 2 . {\displaystyle {\tfrac {1+{\sqrt {5}}}{2}}.\,}

Since the sides of this right triangle are in geometric progression, this is the Kepler triangle.

Thales' theorem

Median of a right angle of a triangle

Thales' theorem states that if A is any point of the circle with diameter BC (except B or C themselves) ABC is a right triangle where A is the right angle. The converse states that if a right triangle is inscribed in a circle then the hypotenuse will be a diameter of the circle. A corollary is that the length of the hypotenuse is twice the distance from the right angle vertex to the midpoint of the hypotenuse. Also, the center of the circle that circumscribes a right triangle is the midpoint of the hypotenuse and its radius is one half the length of the hypotenuse.

Medians

The following formulas hold for the medians of a right triangle:

m a 2 + m b 2 = 5 m c 2 = 5 4 c 2 . {\displaystyle m_{a}^{2}+m_{b}^{2}=5m_{c}^{2}={\frac {5}{4}}c^{2}.}

The median on the hypotenuse of a right triangle divides the triangle into two isosceles triangles, because the median equals one-half the hypotenuse.

The medians ma and mb from the legs satisfy[8]: p.136, #3110

4 c 4 + 9 a 2 b 2 = 16 m a 2 m b 2 . {\displaystyle 4c^{4}+9a^{2}b^{2}=16m_{a}^{2}m_{b}^{2}.}

Euler line

In a right triangle, the Euler line contains the median on the hypotenuse—that is, it goes through both the right-angled vertex and the midpoint of the side opposite that vertex. This is because the right triangle's orthocenter, the intersection of its altitudes, falls on the right-angled vertex while its circumcenter, the intersection of its perpendicular bisectors of sides, falls on the midpoint of the hypotenuse.

Inequalities

In any right triangle the diameter of the incircle is less than half the hypotenuse, and more strongly it is less than or equal to the hypotenuse times $( 2 − 1 ) . {\displaystyle ({\sqrt {2}}-1).}$

[16]: p.281

In a right triangle with legs a, b and hypotenuse c,

c ≥ 2 2 ( a + b ) {\displaystyle c\geq {\frac {\sqrt {2}}{2}}(a+b)}

with equality only in the isosceles case.[16]: p.282, p.358

If the altitude from the hypotenuse is denoted hc, then

h c ≤ 2 4 ( a + b ) {\displaystyle h_{c}\leq {\frac {\sqrt {2}}{4}}(a+b)}

with equality only in the isosceles case.[16]: p.282

Other properties

If segments of lengths p and q emanating from vertex C trisect the hypotenuse into segments of length c/3, then[4]: pp. 216–217

p 2 + q 2 = 5 ( c 3 ) 2 . {\displaystyle p^{2}+q^{2}=5\left({\frac {c}{3}}\right)^{2}.}

The right triangle is the only triangle having two, rather than one or three, distinct inscribed squares.[17]

Given h > k. Let h and k be the sides of the two inscribed squares in a right triangle with hypotenuse c. Then

1 c 2 + 1 h 2 = 1 k 2 . {\displaystyle {\frac {1}{c^{2}}}+{\frac {1}{h^{2}}}={\frac {1}{k^{2}}}.}

These sides and the incircle radius r are related by a similar formula:

1 r = − 1 c + 1 h + 1 k . {\displaystyle \displaystyle {\frac {1}{r}}=-{\frac {1}{c}}+{\frac {1}{h}}+{\frac {1}{k}}.}

The perimeter of a right triangle equals the sum of the radii of the incircle and the three excircles:

a + b + c = r + r a + r b + r c . {\displaystyle a+b+c=r+r_{a}+r_{b}+r_{c}.}

References

^ Howard, G.S.; Bettesworth, J.; Boswell, H.; Stonehouse, F.; Hogg, A. (1788). THE NEW ROYAL CYCLOPAEDIA... Retrieved 2022-03-07.
^ Daniel Parrochia (24 July 2018). Mathematics and Philosophy. John Wiley & Sons. pp. 72–. ISBN 978-1-78630-209-0.
^ Di Domenico, Angelo S., "A property of triangles involving area", Mathematical Gazette 87, July 2003, pp. 323-324.
^ a b Posamentier, Alfred S., and Salkind, Charles T. Challenging Problems in Geometry, Dover, 1996.
^ Wentworth p. 156
^ Voles, Roger, "Integer solutions of $a − 2 + b − 2 = d − 2 {\displaystyle a^{-2}+b^{-2}=d^{-2}}$ ," Mathematical Gazette 83, July 1999, 269–271.
^ Richinick, Jennifer, "The upside-down Pythagorean Theorem," Mathematical Gazette 92, July 2008, 313–317.
^ a b c d e Inequalities proposed in “Crux Mathematicorum”, [1].
^ "Triangle right iff s = 2R + r, Art of problem solving, 2011". Archived from the original on 2014-04-28. Retrieved 2012-01-02.
^ a b c d Andreescu, Titu and Andrica, Dorian, "Complex Numbers from A to...Z", Birkhäuser, 2006, pp. 109-110.
^ "Properties of Right Triangles". Archived from the original on 2011-12-31. Retrieved 2012-02-15.
^ a b c CTK Wiki Math, A Variant of the Pythagorean Theorem, 2011, [2] Archived 2013-08-05 at the Wayback Machine.
^ Darvasi, Gyula (March 2005), "Converse of a Property of Right Triangles", The Mathematical Gazette, 89 (514): 72–76, doi:10.1017/S0025557200176806, S2CID 125992270.
^ Bell, Amy (2006), "Hansen's Right Triangle Theorem, Its Converse and a Generalization" (PDF), Forum Geometricorum, 6: 335–342, archived (PDF) from the original on 2008-07-25
^ Di Domenico, A., "The golden ratio — the right triangle — and the arithmetic, geometric, and harmonic means," Mathematical Gazette 89, July 2005, 261. Also Mitchell, Douglas W., "Feedback on 89.41", vol 90, March 2006, 153-154.
^ a b c Posamentier, Alfred S., and Lehmann, Ingmar. The Secrets of Triangles. Prometheus Books, 2012.
^ Bailey, Herbert, and DeTemple, Duane, "Squares inscribed in angles and triangles", Mathematics Magazine 71(4), 1998, 278-284.