geometry
posted by Maria on .
How do I find the length of the hypotenuse of a right triangle with only one leg length and all angle measurements

Given a leg length "x" and all angles, the hypotenuse is "x" divided by the cosine of the angle adjacent to the given side or "x" divided by the sine of the angle opposite the given side.
You might find the following of some interest.
1There exists a primitive Pythagorean Triple with every odd number as a leg.
2There exists a primitive Pythagorean Triple with every even number as a leg.
3There exists a primitive Pythagorean Triple with every number of the form 4n + 1 as a hypotenuse.
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Given an integer side of a Pythagorean triangle, find the other side and the hypotenuse.
Given the number is odd  Resolve the number into any 2 of its factors, equating the larger to m + n and the smaller to m  n. Solve for m and n and substitute into x = k(m^2  n^2), y = k(2mn), and z = k(m^2 + n^2) where k = 1 for primitive triangles.
Examples:
Given x = 7, a prime. The only 2 factors of 7 are 1 and 7. Therefore, m + n = 7 and m  n = 1. Adding, 2m = 8 making m = 4 and n = 3. Thus, x = 4^2  3^2 = 7, y = 2(4)3 = 24 and z = 4^2 + 3^2 = 25.
Given x = 27, a composite. The factor pairs of 27 are 127 and 39. First, m + n = 27 and m  n = 1. Adding, 2m = 28 making m = 14 and n = 13. Then, x = 14^2  13^2 = 27, y = 2(14)13 = 364 and z = 14^2 + 13^2 = 365.
Also, m + n = 9 and m  n = 3. Adding, 2m = 12 making m = 6 and n = 3. Then, x = 6^2  3^2 = 27, y = 2(6)3 = 36 and z = 6^2 + 3^2 = 45.
An alternate way of finding one of the solutions is to square the odd number giving you 27^2 = 729. 729 = 364 + 365. One solution then becomes 27^2 + 364^2 = 365^2.
Given the number is even  equate the number to 2mn and select m and n to be any two integers which will result in the product 2mn being equal to A.
Examples:
Given y = 14 14 = 2mn making mn = 7. Only 7 and 1 can satisfy this example making m = 7 and n = 1. Then, x = 7^2  1^2 = 48, y = 2(7)1 = 14 and z = 7^2 + 1^2 = 50. Recognize that the 144850 triple is a multiple of the primitive 72425 triple with k = 2.
Given y = 24 24 = 2mn making mn = 12. The factor pairs of 121, 62 and 43 fit this requirement for m and n. Then, x = 12^2  1^2 = 143, y = 2(12)1 = 24 and z = 12^2 + 1^2 = 145.
Also, x = 6^2  2^2 = 32, y = 2(6)2 = 24 and z = 6^2 + 2^2 = 40.
Lastly, x = 4^2  3^2 = 7, y = 2(4)3 = 24 and z = 4^2 + 3^2 = 25.
Note that the first and third triples are primitive while the second is a multiple of the primitive 345 triple with k = 5.
If the given number happens to be the product of two consecutive odd or even numbers.
Given x = 5183. Square root of 5183 = 71.993. 5183 = 71 x 73.
1/71 + 1/73 = 144/5183.
Then, 144^2 = 5183^2 = 5185^2.
Given x = 1368. Square root of 1368 = 36.986. 1368 = 36 x 38.
1/36 + 1/38 = 74/1368.
Then, 74^2 + 1368^2 = 1370^2.
Given any prime number or odd square as a side:
If x is the given prime number side, the other side and the hypotenuse derive from y = [(x^2)/2  .5] and z = [x^2)/2 + .5]. Another way of stating it is that the other side and the hypotenuse are the consecutive integers summing to x^2.
Examples:
Given x = 5. x^2 = 25 making y = 12 ans z = 13.
Given x = 7. x^2 = 49 making y = 24 and z = 25.
Given x = 167. x^2 = 27,889 making y = 13,944 and z = 13,945.
Given x = 9. x^2 = 81 making y = 40 and z = 41.
Given x = 25. x^2 = 625 making y = 312 and z = 313.
In the cases where the odd square has more than 3 factors, there are more answers derivable from the method presented earlier.
Given x = 81.
x^2 = 6561 making y = 3280 and z = 3281.
With factors of 1, 3, 9, 27 and 81, from (m + n) = 81 ans (m  n) = 1 we derive m = 41 and n = 40 leading to x = 81, y = 3280 and z = 3281.
Similarly, from (m + n) = 27 and (m  n) = 3 we derive m = 15 and n = 12 leading to x = 81, y = 360 and z = 369.
Given x = 225
x^2 = 50,625 making y = 25,312 and z = 25,313.
With factors of 1, 3, 5, 9, 15, 25, 45, 75 and 225, from (m + n) = 225 and (m  n) = 1 we derive m = 113 and n = 112 leading to x = 225, y = 25,312 and z = 25,313.
Similarly, from the other factor pairings, we derive three other triples of x = 225, y = 2808 and z = 2817, x = 225, y = 1000 and z = 1025 and x = 225, y = 272 and z = 545.
Given an integer or integer hypotenuse of a Pythagorean triangle, find the two sides.
Every number, prime or composite, of the form 4n + 1 is expressable as the sum of two relatively prime squares in one way and is the hypotenuse of a primitive Pythagorean triangle. Any composite number whose factors are 2 and primes of the form 4n + 1 is similarly expressable as the sum of two relatively prime squares in one way and is the hypotenuse of a primitive Pythagorean triangle. Both the hypotenuse and its square are each the sum of two squares. A power of the prime or double either the prime or power can also be expressed as the sum of two squares.
No number, prime or composite, of the form 4n  1 can be expressed as the sum of two squares or be the hypotenuse of a primitive Pythagorean triangle.
For composite numbers to be expressed as the sum of two squares in more than one way, they must contain a certain number of primes of the form 4n + 1.
...A composite number having only one prime factor of the form 4n + 1 can be expressed as the sum of two squares in one way. The same holds true if there are other factors of 2 or its powers.
...A composite number having two different prime factors of the form 4n + 1 can be expressed as the sum of two squares in two ways. The same holds true if there are other factors of 2 or its powers.
...A composite number having three different prime factors of the form 4n + 1 can be expressed as the sum of two squares in 4 ways. The same holds true if there are other factors of 2 or its powers.
...A composite number having four different prime factors of the form 4n + 1 can be expressed as the sum of two squares in 8 ways. The same holds true if there are other factors of 2 or its powers.
...In general, a composite number having k different prime factors of the form 4n + 1 can be expressed as the sum of two squares in 2^(k1) ways. The same holds true if there are other factors of 2 or its powers.
A composite number having an even number of prime factors of the form 4n  1 and one or more prime factors of the form 4n + 1 can be expressed as the sum of two squares, but in only one way. The inclusion of the factor of 2, or any power of 2, will not alter the ability of the number to be expressed as the sum of two Squares nor increase the number of ways that it can be so expressed.
If N is a perfect square and the sqrt(N) = z is of the form 4n + 1 (an odd number), z is the hypotenuse of a primitive or nonprimitive Pythagorean triangle and N is expressable as the sum of two Squares.
If N is a perfect square and the sqrt(N) = z is even, N may be the hypotenuse of a nonprimitive Pythagorean triangle.
A hypotenuse of a primitive, or nonprimitive, Pythagorean triangle is, by definition, of the form of k(m^2+n^2).
Examples:
Given 4177
Are there integers x and y that satisfy x^2 + y^2 = 4177?
Are there integers x and y that satisfy x^2 + y^2 = z^2 = 4177^2?
What values of x and y satisfy x^2 + y^2 = 4177?
SInce N ends in 7, it is not a square. Taking the square root clearly verifies this.
Since N ends in 7, it might be a prime.
Dividing by all primes less than sqrt(4177) = 64.62 will verify whether it is prime. This requires 18 calculations.
The calculations verify that N = 4177 is prime.
Being of the form 4n + 1, it is expressable as the sum of 2 squares in one way only.
The minimum possible value of "x" is sqrt[4177/2] = 45.7.
The maximum possible value of "x" is sqrt[4177] = 64.62.
Therefore, 45 < x < 65.
SInce 4177  x^2 = y^2, and the last two digits of the two squares are required to end in 00, 01, 04, 09, 16, 21, 24, 25, 29, 36, 41, 44, 49, 56, 61, 64, 69, 76, 81, 84, 89, or 96, inspection leads us to the fact that "x" must end in one of the following:
x^2 ending.....01.....16.....21.....36.....41.....56.....61.....76.....81.....96
y^2 ending....76......61.....56.....41.....36.....21.....16.....01.....96.....81
Note that the 5th through 8th pairs are simply the reverse of the 1st through 4th pairs and the 10th is simply the opposite of the 9th.
In reality, we need only deal with 5 possible pairings of numbers as the other 5 are merely opposites.
x^2 ending.....01.....16.....21.....36.....81
y^2 ending.......76......61....56.....41.....96
The only numbers within our working range of 46 to 64 that can have any of these 5 endinga when squared are 49, 51, 59 and 61 for those ending in a 1 and 46, 54, 56 and 64 for those ending in a 6. Eight quick calculations, depending on which end of the possible answers we start with, leads us to x = 64.
With x = 64, 64^2 = 4096 resulting in y^2 = 81 or y = 9 making 9^2 + 64^2 = 4177.
We know that primitive Pythagorean triangles derive from x = m^2  n^2, y = 2mn and z = m^2 + n^2, m greater than n, one odd one even. The "x" and "y" values we derived thus become the "m" and "n" values.
Our primitive Pythagorean Triple then becomes x = 64^2  9^2 = 4015, y = 2(64)9 = 1152 and z = 64^2 + 9^2 = 4177, the triple of 115240154177 whereby 1152^2 + 4105^2 = 4177^2.
Given 5429
Are there integers that satisfy x^2 + y^2 = 5429?
Are there integers x and y that satisfy x^2 + y^2 = 5429^2?
What values of x and y satisfy x^2 + y^2 = 5429?
SInce N ends in 9, it might be a square.
The sum of the digits is not 1, 4, 7 or 9 making it not a square.
SInce N ends in 9, it might be a prime.
Dividing by all primes less than sqrt(5429) = 73.68 will verify whether it is prime. This requires 21 calculations.
The calculations verify that N = 5429 is composite having prime factors of 61 and 89, both of which are of the form 4n + 1.
Therefore, 5429 is the sum of two squares in two different ways.
The minimum possible value of "x" is sqrt[5429/2] = 52.1.
The maximum possible value of "x" is sqrt[5429] = 73.68
Therefore, 53 < x < 74.
SInce 5429  x^2 = y^2, and the last two digits of the two squares are required to end in 00, 01, 04, 09, 16, 21, 24, 25, 29, 36, 41, 44, 49, 56, 61, 64, 69, 76, 81, 84, 89, or 96, inspection leads us to the fact that "x" must end in one of the following:
x^2 ending.....00.....04.....25.....29
y^2 ending.......29.....25.....04.....00
The numbers within our range of 54 to 73 that can end in any of these 4 endings when squared are 55, 57, 60, 62, 63, 65, 67, 70, 72 and 73 . Ten quick calculations leads us to x = 70 and 73.
With x = 70, x^2 = 4900 resulting in y^2 = 529 making y = 23.
With x = 73, x^2 = 5329 resulting is y^2 = 100 making y = 10.
The first solution is found to be 70^2 + 23^2 = 4900 + 529 = 5429.
The second solution is found to be 73^2 + 10^2 = 5329 + 100 = 5429.
We also know that N = 5429 is the hypotenuse of 2 primitive Pythagorean triangles.
We know that primitive Pythagorean triangles derive from a = m^2  n^2, b = 2mn and c = m^2 + n^2, m greater than n, one odd one even.
The "x" and "y" values we derived become the "m" and "n" values.
Our first primitive Pythagorean triangle then becomes x = 70^2  23^2 = 4371, y = 2(70)23 = 3220 and z = 70^2 + 23^2 = 5429, the triple of 322043715429 for 3220^2 + 4371^2 = 5429^2.
Our second primitive Pythagorean triangle then becomes x = 73^2  10^2 = 5229, y = 2(73)10 = 1460 and z = 73^2 + 10^2 = 5429, the triple of 146052295429 for 1460^2 + 5229^2 = 5429^2.
Given 5329
Are there integers that satisfy x^2 + y^2 = 5329?
Are there integers x and y that satisfy x^2 + y^2 = 5329^2?
Since N ends in 9, the sum of its digits is 1, and the last 2 digits are 29, there is a strong possibility that 5329 is a square.
Taking the square root leads us to sqrt5329 = 73, a prime.
What values of m and n would lead to a primitive hypotenuse of 73?
We can work out x^2 + y^2 = 73^2 by inspection by noting that (8^2 + 3^2) = 73.
Thus, the primitive triangle with hypotenuse of 73 has sides of x = 8^2  3^2 = 55 and y = 2(8)3 = 48 forming the triple of 485573 where 48^2 + 55^2 = 73^2.
Working out the values of x and y satisfying x^2 + y^2 = 5329 the long way:
The minimum possible value of "x" is sqrt[5329/2] = 51.61
The maximum possible value of "x" is sqrt[5329] = 73.
Therefore, 51 < x < 74.
SInce 5329  x^2 = y^2, and the last two digits of the two squares are required to end in 00, 01, 04, 09, 16, 21, 24, 25, 29, 36, 41, 44, 49, 56, 61, 64, 69, 76, 81, 84, 89, or 96, inspection leads us to the fact that "x" must end in one of the following:
x^2 ending.....00.....04.....25.....29
y^2 ending.......29.....25.....04.....00
The numbers within our range of 52 to 73 that can end in any of these 4 endings when squared are 53, 55, 57, 60, 62, 63, 65, 67, 70, 72 and 73 . Ten quick calculations leads us to x = 55.
With x = 55, x^2 = 3025 resulting in y^2 = 2304 making y = 48.
Therefore, 55^2 + 48^2 = 5329.
Our Pythagorean triple is therefore x = 8^2  3^2 = 55, y = 2(8)3) = 48 and z = 8^2 + 3^2 = 73 from which we obtain 55^2 + 48^2 = 73^2 = 5329.
Given 6409
Are there integers x and y that satisfy x^2 + y^2 = 6409^2?
What values of x and y satisfy x^2 + y^2 = 6409?
SInce N ends in 9, it might be a square.
Since the sum of the digits is 1, it still might be a square.
SInce N ends in 9, it might be a square or a prime.
Dividing by all primes less than sqrt(6409) = 80.05 will verify whether it is prime. This requires 22 calculations.
The calculations verify that N = 6409 is not a prime with prime factors of 13, 17 and 29, all of which are of the form 4n + 1.
Therefore, 6409 is the sum of 2 squares in 2^(31) = 4 different ways.
The minimum possible value of "x" is sqrt[6409/2] = 56.6.
The maximum possible value of "x" is sqrt[6408] = 80.05
Therefore, 56 < x < 81.
Starting with x = 80 and working our way through the lower values, we quickly reach the 4 pairs of values where, ...with x = 80, x^2 = 6400 requiring that y^2 = 9 resulting in y = 3.
...with x = 75, x^2 = 5625 requiring that y^2 = 784 resulting in y = 28
...with x = 72, x^2 = 5184 requiring that y^2 = 1225 resulting in y = 35
...with x = 60, x^2 = 3600 requiring that y^2 = 2809 resulting in y = 53.
The four solutions are therefore 80^2 + 3^2 = 6400 + 9 = 6409
.............................................75^2 + 28^2 = 5625 + 784 = 6409
.............................................72^2 + 35^2 = 5184 + 1225 = 6409
.............................................60^2.+ 53^2 = 3600 + 2809 = 6409..
We also know that N = 6409, as an odd composite, is the hypotenuse of 4 primitive Pythagorean triangle.
Primitive Pythagorean triangles derive from x = k(m^2  n^2), y = 2kmn and z = k(m^2 + n^2), m greater than n, one odd one even, k = 1 for primitiveness.
The "x" and "y" values we derived become the "m" and "n" values and our primitive Pythagorean triangles become
x = 80^2  3^2 = 6391, y = 2(80)3 = 480 and z = 80^2 + 3^2 = 6409
x = 75^2  28^2 = 4841, y = 2(75)28 = 4200 and z = 75^2 + 28^2 = 6409
x = 72^2  35^2 = 3959, y = 2(72)35 = 5040 and z = 72^2 + 35^2 = 6409 and
x = 60^2  53^2 = 791, y = 2(60)53 = 6360 and z = 60^2 + 53^2 = 6409..
FInding Pythagorean triangles with consecutive legs.
We are already familiar with the 345 triangle, the only triple with all three legs consecutive.
If the two legs are to be consecutive, they must satisfy (m^2  n^2) + 1 = 2mn or (m^2  n^2) = 2mn + 1. Adding n^2 to both sides results in (m  n)^2 = 2n^2 + 1 and (m  n)^2 = 2n^2  1. Thus, (m  n)^2 = 2n^2 +/1. An equation of this form, x^2  Dy^2 = +/1 is typically referred to as a Pell equation even though it was Fermat who really discovered it. Solutions to Pell equations do not come easy and the method is covered in many books on number theory and possibly elsewhere in the Knowledge Database. Here, we will simply make use of the tried and true trial and error method to see how many solutions we can find within a reasonable range of values.
What we seek are values of 2n^2 +/ 1 that are squares and therefore equal to (m  n)^2. Lets see how many we can find.
.........(mn)^2.....(mn)^2
.n......2n^2+1......2n^21......(m  n)......m.......x.......y.......z
.1..........3..............1..............1..........2........3.......4.......5
.2..........9..............7..............3..........5.......21.....20.....29
.3.........19............17
.4.........33............31
.5.........51............49.............7..........12....119....120...169
.6.........73............71
.7.........99............97
.8.......129...........127
.9.......163...........161
10......201...........199
11......243...........241
12......289...........287............17.........19....697....696...985
13......339...........337
14......393...........391
15......451...........449
Note the enlarged underlined values as being the perfect squares which lead to the 4 solutions shown. The list goes on but, sufficeth to say, there are many more.
Consecutive legs can also be derived from the "m" and "n" values derived from the general expression ni = 2(n(p1) + m(p2) and mi = 2m(p1) + n(p1). The subscript "p' indicates the "previous" value. The values m(p1) and n(p1) are the previous "m" and "n" values while m(p2) and n(p2) are the values previous to m(p1) and n(p1). Consider the starting point of the 345 triple which derives from "m" and "n" values of 2 and 1. The next triple that will produce a triple with both legs consecutive comes from the values of n = 2(1) + 0 = 2 and m = 2(2) + 1 = 5 which produces the 212029 triple. The next will derive from n = 2(2) + 1 = 5 and m = 2(5) + 2 = 12 producing the 119120169 triple. See how many more you can create and see if you can determine another interesting relationship between the results.
Finding Pythagorean triangles with a leg and hypotenuse consecutive.
If a leg and hypotenuse are to be consecutive, they must satisfy (m^2 + n^2) = (m^2  n^2) + 1 or (m^2 + n^2) = 2mn + 1. The first reduces to 2n^2 = 1 meaning there is no rational solution. The second leads to m^2  2mn + n^2 = 1 or (m  n)^2 = 1 or (m  n) = 1 making m = n + 1. Therefore, all Pythagorean triangles derived from consecutive values of m and n have a leg and the hypotenuse consecutive integers.
Examples:
n......m.......x.......y.......z
1......2........3.......4.......5
2......3........5......12.....13
3......4........7......24.....25
4......5........9......40.....41
5......6.......11.....60.....61
Finding Pythagorean triangles having equal areas and perimeters.
The three expressions for deriving the integer sides of a Pythagorean triangle, Pythagorean Triples, are x = m^2  n^2, y = 2mn, and z = m^2 + n^2. The area can be expressed by A = 2mn(m^2  n^2)/2. The perimeter can be expressed by P = m^2  n^2 + 2mn + m^2 + n^2. Equating the two and simplifying we get m^2  (2/n)m  (n^2 + 2) = 0. Solving for m using the quadratic formula, we derive m = [(2/n) +/ sqrt[(4/n^2) + 4n^2 + 8]]/2. The 4/n^2 allows only n = 1 or 2 in order to result in integer answers which results in m = 3 in both cases. Therefore, there are only 2 Pythagorean triangles whose areas and perimeters are equal. One derived from m = 3 and n = 1 and one from m = 3 and n = 2 which turn out to be the 6810 and the 51213 triangles.
From another viewpoint considering all right triangles:
Consider the right triangle with sides a, b, and c. The area A = ab/2 and the perimeter p = a + b + c. With a right triangle, c = sqrt(a^2 + b^2) making ab/2 = a + b + sqrt(a^2 + b^2). This leads to ab/2  a  b = sqrt(a^2 + b^2). Squaring both sides, collecting and simplifying leads to ab/4 +2 a b = 0 which leads to b = (a  2)/(a/4  1) or b = 4(a  2)/(a  4). Thus, there is a "b" for every value of "a" greater than 4. Values of a = 5 and 6 lead us to the same integer triangles already identified, 51213 and 6810. Values beyond a = 6 produce an infinite number of right triangle answers where at least one side is an integer and at most two.
Of how many primitive Pythagorean triangles may N be a leg?
If a number N is made up from "n" primes and their powers, it is the leg of 2^(n  1) primitive Pythagreaon
triangles.
Example: For N = 2700, its 3 factors, 2^2(3^3)5^2, give rise to 2^(31) = 4 primitive triangles. (If N is even but not
divisible by 4, there are no solutions, as there cannot be for any other number of the form 4x + 2.)
Of how many primitive and nonprimitive Pythagorean triangles may N be a leg?
Given N = 2^(ao)(p1^(a1))(p2^(a2))....pn^(an), the number of triangles N can be a leg of is given by
Legs = [(2(ao)  1)(2(a1) + 1)(2(a2) + 1)....(2(an) + 1)  1]/2.
Example: For N = 2700, where N = 2^2(3^3)5^2, the number of possible triangles with N = 2700 is
L = [(2(2)  1)(2(3) + 1)(2(2) + 1) 1]/2 = 52.
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Having explored Pythagorean triple triangles, you might be even more amazed at the unique properties of nonPythagorean, or scalene, integer triangles, better known as Heronian triangles and their connection to Pythagorean triple triangles. A Heronian, or arithmetical, triangle is one whose sides, area, and at least one altitude, are all integers. Some mathematicians refer to Heronian triangles as rational triangles. You are no doubt asking yourself exactly what do Heronian triangles and Pythagorean triangles have in common?
Since the sides of a Heronian triangle are integral, the integer altitude from a vertex to the opposite side forms two integer right triangles whose sum, or difference, equal the given triangle, or a smaller Heronian triangle, respectively. Knowing this, it is possible to form any number of Heronian triangles, which are not right triangles, by combining two integral right triangles of different shapes having a common leg. For instance, given the 51213 and 91215 right triangles, we can join them at the common 12 leg and create the 131415 and 13415 triangles, each with altitude 12.
Using the 51213 and 121620 right triangles, with the 12 length sides coincident, the Heronian triangle thus formed has sides of 132021, altitude 12, and area 126. Rotating the smaller 51213 triangle about the common altitude, inside the larger 121620 triangle, we derive another Heronian triangle with sides 111320, altitude 12, and area 66.
Using the 345 and 51213 primitive right triangles, we can combine the primitive 51213 with the nonprimitive derivatives 91215 and 121620, and the nonprimitive derivative 52025 with the nonprimatives 153639 and 204852, to create eight different Heronian triangles.
Similarly, isosceles Heronian triangles can be formed from either of the given Heronian triangle by rotating the triangle about a common altitude. For example, the 51213 triangle forms 101313 or 241313 isosceles Heronian triangles with areas of 60.
Given two primitive right triangles, it is possible to form proportionally larger similar triangles having either leg of one equal to either leg of the other. We can then form Heronian triangles by bringing the equal legs coincident with one another, with the unequal legs on a common perpendicular to the common leg, either on the same side of it or on opposite sides of the common leg.
Take the 51213 and 72425 primitive triangles and create the proportionally triangles of 358491, 35120125, 84288300 and 120288312 triangles. Combining these in all possible ways leads to the Heronian triangles of
913612535, 9120412535, 30036312288, 300204312288, 30012525384, 3009132384, 312125253120 and 312125323120. The last number in each set of four numbers is the integer altitude for that particular triangle. In addition to these scalene Heronian triangles, the following isosceles Heronian triangles are also derivable; 312312576, 300300576, 12512570, 919170, 312312240, 300300168, 125125240 and 9191168.