Finding the adjacent of a right angle triangle with given height.
$begingroup$
Given the following conditions:
- A right angled triangle
- Height is 13
- Angle a & b are not equal
- X and Y are not prime numbers.
Is there any math formula to find X,Y, A & B?
Thank you.
trigonometry
edited Dec 16 '18 at 11:09
N. F. Taussig
44.2k93356
asked Dec 16 '18 at 9:03
user549534user549534
61
$endgroup$
add a comment |
$begingroup$
Given the following conditions:
- A right angled triangle
- Height is 13
- Angle a & b are not equal
- X and Y are not prime numbers.
Is there any math formula to find X,Y, A & B?
Thank you.
trigonometry
edited Dec 16 '18 at 11:09
N. F. Taussig
44.2k93356
asked Dec 16 '18 at 9:03
user549534user549534
61
$endgroup$
$begingroup$
Do you want all sides positive integers?
$endgroup$
– coffeemath
Dec 16 '18 at 11:03
$begingroup$
user549534 -- Have you had a chance to look at my answer below? If so any questions?
$endgroup$
– coffeemath
Dec 17 '18 at 0:41
add a comment |
$begingroup$
Given the following conditions:
- A right angled triangle
- Height is 13
- Angle a & b are not equal
- X and Y are not prime numbers.
Is there any math formula to find X,Y, A & B?
Thank you.
trigonometry
edited Dec 16 '18 at 11:09
N. F. Taussig
44.2k93356
asked Dec 16 '18 at 9:03
user549534user549534
61
$endgroup$
Given the following conditions:
- A right angled triangle
- Height is 13
- Angle a & b are not equal
- X and Y are not prime numbers.
Is there any math formula to find X,Y, A & B?
Thank you.
trigonometry
trigonometry
edited Dec 16 '18 at 11:09
N. F. Taussig
44.2k93356
asked Dec 16 '18 at 9:03
user549534user549534
61
edited Dec 16 '18 at 11:09
N. F. Taussig
44.2k93356
asked Dec 16 '18 at 9:03
user549534user549534
61
edited Dec 16 '18 at 11:09
N. F. Taussig
44.2k93356
edited Dec 16 '18 at 11:09
N. F. Taussig
44.2k93356
edited Dec 16 '18 at 11:09
N. F. Taussig
44.2k93356
44.2k93356
asked Dec 16 '18 at 9:03
user549534user549534
61
asked Dec 16 '18 at 9:03
user549534user549534
61
asked Dec 16 '18 at 9:03
user549534user549534
61
61
$begingroup$
Do you want all sides positive integers?
$endgroup$
– coffeemath
Dec 16 '18 at 11:03
$begingroup$
user549534 -- Have you had a chance to look at my answer below? If so any questions?
$endgroup$
– coffeemath
Dec 17 '18 at 0:41
add a comment |
$begingroup$
Do you want all sides positive integers?
$endgroup$
– coffeemath
Dec 16 '18 at 11:03
$begingroup$
user549534 -- Have you had a chance to look at my answer below? If so any questions?
$endgroup$
– coffeemath
Dec 17 '18 at 0:41
$begingroup$
Do you want all sides positive integers?
$endgroup$
– coffeemath
Dec 16 '18 at 11:03
$begingroup$
Do you want all sides positive integers?
$endgroup$
– coffeemath
Dec 16 '18 at 11:03
$begingroup$
user549534 -- Have you had a chance to look at my answer below? If so any questions?
$endgroup$
– coffeemath
Dec 17 '18 at 0:41
$begingroup$
user549534 -- Have you had a chance to look at my answer below? If so any questions?
$endgroup$
– coffeemath
Dec 17 '18 at 0:41
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your triangle will be a primitive Pythagorean one since one side is $13$ and no right triangle has all three sides equal. It's known that the sides of such are $p^2-q^2,2pq,p^2+q^2$ [last being hypotenuse] where $p,q$ have gcd $1$ and $p>q.$
$13$ is odd so it must be $p^2-q^2=(p-q)(p+q)$ the only way this is $13$ is if $p-q=1,p+q=13$ leading to $p=7,q=6.$ Then the other leg is $2pq=84$ and hypotenuse $85.$
Using a rough approx. in calculator gives $A=8.8$ and $B=81.2$ [degrees] Used arctan for these and rounded.
answered Dec 16 '18 at 11:48
coffeemathcoffeemath
2,8451415
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
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votes
$begingroup$
Your triangle will be a primitive Pythagorean one since one side is $13$ and no right triangle has all three sides equal. It's known that the sides of such are $p^2-q^2,2pq,p^2+q^2$ [last being hypotenuse] where $p,q$ have gcd $1$ and $p>q.$
$13$ is odd so it must be $p^2-q^2=(p-q)(p+q)$ the only way this is $13$ is if $p-q=1,p+q=13$ leading to $p=7,q=6.$ Then the other leg is $2pq=84$ and hypotenuse $85.$
Using a rough approx. in calculator gives $A=8.8$ and $B=81.2$ [degrees] Used arctan for these and rounded.
answered Dec 16 '18 at 11:48
coffeemathcoffeemath
2,8451415
$endgroup$
add a comment |
$begingroup$
Your triangle will be a primitive Pythagorean one since one side is $13$ and no right triangle has all three sides equal. It's known that the sides of such are $p^2-q^2,2pq,p^2+q^2$ [last being hypotenuse] where $p,q$ have gcd $1$ and $p>q.$
$13$ is odd so it must be $p^2-q^2=(p-q)(p+q)$ the only way this is $13$ is if $p-q=1,p+q=13$ leading to $p=7,q=6.$ Then the other leg is $2pq=84$ and hypotenuse $85.$
Using a rough approx. in calculator gives $A=8.8$ and $B=81.2$ [degrees] Used arctan for these and rounded.
answered Dec 16 '18 at 11:48
coffeemathcoffeemath
2,8451415
$endgroup$
add a comment |
$begingroup$
Your triangle will be a primitive Pythagorean one since one side is $13$ and no right triangle has all three sides equal. It's known that the sides of such are $p^2-q^2,2pq,p^2+q^2$ [last being hypotenuse] where $p,q$ have gcd $1$ and $p>q.$
$13$ is odd so it must be $p^2-q^2=(p-q)(p+q)$ the only way this is $13$ is if $p-q=1,p+q=13$ leading to $p=7,q=6.$ Then the other leg is $2pq=84$ and hypotenuse $85.$
Using a rough approx. in calculator gives $A=8.8$ and $B=81.2$ [degrees] Used arctan for these and rounded.
answered Dec 16 '18 at 11:48
coffeemathcoffeemath
2,8451415
$endgroup$
Your triangle will be a primitive Pythagorean one since one side is $13$ and no right triangle has all three sides equal. It's known that the sides of such are $p^2-q^2,2pq,p^2+q^2$ [last being hypotenuse] where $p,q$ have gcd $1$ and $p>q.$
$13$ is odd so it must be $p^2-q^2=(p-q)(p+q)$ the only way this is $13$ is if $p-q=1,p+q=13$ leading to $p=7,q=6.$ Then the other leg is $2pq=84$ and hypotenuse $85.$
Using a rough approx. in calculator gives $A=8.8$ and $B=81.2$ [degrees] Used arctan for these and rounded.
answered Dec 16 '18 at 11:48
coffeemathcoffeemath
2,8451415
answered Dec 16 '18 at 11:48
coffeemathcoffeemath
2,8451415
answered Dec 16 '18 at 11:48
coffeemathcoffeemath
2,8451415
answered Dec 16 '18 at 11:48
coffeemathcoffeemath
2,8451415
2,8451415
add a comment |
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$begingroup$
Do you want all sides positive integers?
$endgroup$
– coffeemath
Dec 16 '18 at 11:03
$begingroup$
user549534 -- Have you had a chance to look at my answer below? If so any questions?
$endgroup$
– coffeemath
Dec 17 '18 at 0:41