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#ActualKhatharr

Posted 01 February 2013 - 06:18 AM

Calc class is covering pythagoras and trig, which is pretty familiar, so I'm slightly bored which leads to mischief.
 
If I had a right triangle with a known hypotenuse and one angle is it possible to get the lengths of the other two sides without using cos/csc/sin/sec? In other words only using tan/cot and pythagoras?
 
It seems like it should be possible. With pythagoras I can get the sum of a² and b² and with tan/cot I can get the ratio between a and b. I'm fiddling around on paper trying to do it as a formula but it's not simplifying for me.
 
With sides a,b,c and angles A,B,C where the matching letters are opposites (side opposite angle) and c is the hypotenuse, given A and c:
 
a² + b² = c²
(a*b/b)² + (b*a/a)² = c²
(b*a/b)² + (a*b/a)² = c²
(b*tan(A))² + (a*cot(A))² = c²
b²*tan(A)² + a²*cot(A)² = c²
b²*tan(A)² = c² - a²*cot(A)²
b² = (c² - a²*cot(A)²) / tan(A)²
b = sqrt((c² - a²*cot(A)²) / tan(A)²)
 
So I can solve for b in terms of a or a in terms of b...
 
Eh... Is this possible? (OMG I'm such a nerd...)

Or maybe something like:

a² + b² = c²
a² = c² - b²
a = sqrt(c² - b²)
tan(A) = a/b
b*tan(A) = a
b*tan(A) = sqrt(c² - b²)

Which eliminates a but I don't know if that b² is recoverable...

Okay, so...

Say A = 36.8699° and c = 10
tan(A) = 3/4

We could say that:
a = 3x
b = 4x

(3x)² + (4x)² = 10²
9x² + 16x² = 100
25x² = 100
x² = 100/25
x² = 4
x = 2

a = 6
b = 8

Which is correct, so this is possible with that method, but now I'm wondering how I can express that as a formula...

Oh, I got it...

Say A = 36.8699° and c = 10
tan(A) = 0.75

a = 0.75b

(0.75b)² + b² = 10²
0.5625b² + b² = 100
1.5625b² = 100
b² = 100/1.5625
b² = 64
b = 8
a = 0.75*8 = 6

Schwing! Achievement unlocked.

(tan(A)² * b²) + b² = c²

(tan(A)² + 1) * b² = c²

b² = c² / (tan(A)² + 1)

 

b = c / sqrt(tan(A)² + 1)


a = b*tan(A)

Hope someone else enjoys this.


#9Khatharr

Posted 01 February 2013 - 06:17 AM

<p>Calc class is covering pythagoras and trig, which is pretty familiar, so I'm slightly bored which leads to mischief.<br />
&nbsp;<br />
If I had a right triangle with a known hypotenuse and one angle is it possible to get the lengths of the other two sides without using cos/csc/sin/sec? In other words only using tan/cot and pythagoras?<br />
&nbsp;<br />
It seems like it should be possible. With pythagoras I can get the sum of a&sup2; and b&sup2; and with tan/cot I can get the ratio between a and b. I'm fiddling around on paper trying to do it as a formula but it's not simplifying for me.<br />
&nbsp;<br />
With sides a,b,c and angles A,B,C where the matching letters are opposites (side opposite angle) and c is the hypotenuse, given A and c:<br />
&nbsp;<br />
a&sup2; + b&sup2; = c&sup2;<br />
(a*b/b)&sup2; + (b*a/a)&sup2; = c&sup2;<br />
(b*a/b)&sup2; + (a*b/a)&sup2; = c&sup2;<br />
(b*tan(A))&sup2; + (a*cot(A))&sup2; = c&sup2;<br />
b&sup2;*tan(A)&sup2; + a&sup2;*cot(A)&sup2; = c&sup2;<br />
b&sup2;*tan(A)&sup2; = c&sup2; - a&sup2;*cot(A)&sup2;<br />
b&sup2; = (c&sup2; - a&sup2;*cot(A)&sup2;) / tan(A)&sup2;<br />
b = sqrt((c&sup2; - a&sup2;*cot(A)&sup2;) / tan(A)&sup2;)<br />
&nbsp;<br />
So I can solve for b in terms of a or a in terms of b...<br />
&nbsp;<br />
Eh... Is this possible? (OMG I'm such a nerd... <img alt="wacko.png" src="http://public.gamedev.net//public/style_emoticons/default/wacko.png" title=":wacko:" /> )<br />
<br />
Or maybe something like:<br />
<br />
a&sup2; + b&sup2; = c&sup2;<br />
a&sup2; = c&sup2; - b&sup2;<br />
a = sqrt(c&sup2; - b&sup2;)<br />
tan(A) = a/b<br />
b*tan(A) = a<br />
b*tan(A) = sqrt(c&sup2; - b&sup2;)<br />
<br />
Which eliminates a but I don't know if that b&sup2; is recoverable...<br />
<br />
Okay, so...<br />
<br />
Say A = 36.8699&deg; and c = 10<br />
tan(A) = 3/4<br />
<br />
We could say that:<br />
a = 3x<br />
b = 4x<br />
<br />
(3x)&sup2; + (4x)&sup2; = 10&sup2;<br />
9x&sup2; + 16x&sup2; = 100<br />
25x&sup2; = 100<br />
x&sup2; = 100/25<br />
x&sup2; = 4<br />
x = 2<br />
<br />
a = 6<br />
b = 8<br />
<br />
Which is correct, so this is possible with that method, but now I'm wondering how I can express that as a formula...<br />
<br />
Oh, I got it...<br />
<br />
Say A = 36.8699&deg; and c = 10<br />
tan(A) = 0.75<br />
<br />
a = 0.75b<br />
<br />
(0.75b)&sup2; + b&sup2; = 10&sup2;<br />
0.5625b&sup2; + b&sup2; = 100<br />
1.5625b&sup2; = 100<br />
b&sup2; = 100/1.5625<br />
b&sup2; = 64<br />
b = 8<br />
a = 0.75*8 = 6<br />
<br />
Schwing! Achievement unlocked.<br />
<br />
(tan(A)&sup2; * b&sup2;) + b&sup2; = c&sup2;</p>
<p>(tan(A)&sup2; + 1) * b&sup2; = c&sup2;</p>
<p>b&sup2; = c&sup2; / (tan(A)&sup2; + 1)</p>
<p>&nbsp;</p>
<p>b = c / sqrt(tan(A)&sup2; + 1)</p>
<p><br />
a = b*tan(A)<br />
<br />
Hope someone else enjoys this. <img alt="laugh.png" src="http://public.gamedev.net//public/style_emoticons/default/laugh.png" title=":lol:" /></p>

#8Khatharr

Posted 01 February 2013 - 04:55 AM

Calc class is covering pythagoras and trig, which is pretty familiar, so I'm slightly bored which leads to mischief.
 
If I had a right triangle with a known hypotenuse and one angle is it possible to get the lengths of the other two sides without using cos/csc/sin/sec? In other words only using tan/cot and pythagoras?
 
It seems like it should be possible. With pythagoras I can get the sum of a² and b² and with tan/cot I can get the ratio between a and b. I'm fiddling around on paper trying to do it as a formula but it's not simplifying for me.
 
With sides a,b,c and angles A,B,C where the matching letters are opposites (side opposite angle) and c is the hypotenuse, given A and c:
 
a² + b² = c²
(a*b/b)² + (b*a/a)² = c²
(b*a/b)² + (a*b/a)² = c²
(b*tan(A))² + (a*cot(A))² = c²
b²*tan(A)² + a²*cot(A)² = c²
b²*tan(A)² = c² - a²*cot(A)²
b² = (c² - a²*cot(A)²) / tan(A)²
b = sqrt((c² - a²*cot(A)²) / tan(A)²)
 
So I can solve for b in terms of a or a in terms of b...
 
Eh... Is this possible? (OMG I'm such a nerd... wacko.png )

Or maybe something like:

a² + b² = c²
a² = c² - b²
a = sqrt(c² - b²)
tan(A) = a/b
b*tan(A) = a
b*tan(A) = sqrt(c² - b²)

Which eliminates a but I don't know if that b² is recoverable...

Okay, so...

Say A = 36.8699° and c = 10
tan(A) = 3/4

We could say that:
a = 3x
b = 4x

(3x)² + (4x)² = 10²
9x² + 16x² = 100
25x² = 100
x² = 100/25
x² = 4
x = 2

a = 6
b = 8

Which is correct, so this is possible with that method, but now I'm wondering how I can express that as a formula...

Oh, I got it...

Say A = 36.8699° and c = 10
tan(A) = 0.75

a = 0.75b

(0.75b)² + b² = 10²
0.5625b² + b² = 100
1.5625b² = 100
b² = 100/1.5625
b² = 64
b = 8
a = 0.75*8 = 6

Schwing! Achievement unlocked.

b = c/(1+tan(A))
a = b*tan(A)

Hope someone else enjoys this. laugh.png


#7Khatharr

Posted 01 February 2013 - 04:54 AM

Calc class is covering pythagoras and trig, which is pretty familiar, so I'm slightly bored which leads to mischief.
 
If I had a right triangle with a known hypotenuse and one angle is it possible to get the lengths of the other two sides without using cos/csc/sin/sec? In other words only using tan/cot and pythagoras?
 
It seems like it should be possible. With pythagoras I can get the sum of a² and b² and with tan/cot I can get the ratio between a and b. I'm fiddling around on paper trying to do it as a formula but it's not simplifying for me.
 
With sides a,b,c and angles A,B,C where the matching letters are opposites (side opposite angle) and c is the hypotenuse, given A and c:
 
a² + b² = c²
(a*b/b)² + (b*a/a)² = c²
(b*a/b)² + (a*b/a)² = c²
(b*tan(A))² + (a*cot(A))² = c²
b²*tan(A)² + a²*cot(A)² = c²
b²*tan(A)² = c² - a²*cot(A)²
b² = (c² - a²*cot(A)²) / tan(A)²
b = sqrt((c² - a²*cot(A)²) / tan(A)²)
 
So I can solve for b in terms of a or a in terms of b...
 
Eh... Is this possible? (OMG I'm such a nerd... wacko.png )

Or maybe something like:

a² + b² = c²
a² = c² - b²
a = sqrt(c² - b²)
tan(A) = a/b
b*tan(A) = a
b*tan(A) = sqrt(c² - b²)

Which eliminates a but I don't know if that b² is recoverable...

Okay, so...

Say A = 36.8699° and c = 10
tan(A) = 3/4

We could say that:
a = 3x
b = 4x

(3x)² + (4x)² = 10²
9x² + 16x² = 100
25x² = 100
x² = 100/25
x² = 4
x = 2

a = 6
b = 8

Which is correct, so this is possible with that method, but now I'm wondering how I can express that as a formula...

Oh, I got it...

Say A = 36.8699° and c = 10
tan(A) = 0.75

a = 0.75b

(0.75b)² + b² = 10²
0.5625b² + b² = 100
1.5625b² = 100
b² = 100/1.5625
b² = 64
b = 8
a = 0.75*8 = 6

Schwing! Achievement unlocked.

b = c/(1+tan(A))
a = b*tan(A)

#6Khatharr

Posted 01 February 2013 - 04:50 AM

Calc class is covering pythagoras and trig, which is pretty familiar, so I'm slightly bored which leads to mischief.
 
If I had a right triangle with a known hypotenuse and one angle is it possible to get the lengths of the other two sides without using cos/csc/sin/sec? In other words only using tan/cot and pythagoras?
 
It seems like it should be possible. With pythagoras I can get the sum of a² and b² and with tan/cot I can get the ratio between a and b. I'm fiddling around on paper trying to do it as a formula but it's not simplifying for me.
 
With sides a,b,c and angles A,B,C where the matching letters are opposites (side opposite angle) and c is the hypotenuse, given A and c:
 
a² + b² = c²
(a*b/b)² + (b*a/a)² = c²
(b*a/b)² + (a*b/a)² = c²
(b*tan(A))² + (a*cot(A))² = c²
b²*tan(A)² + a²*cot(A)² = c²
b²*tan(A)² = c² - a²*cot(A)²
b² = (c² - a²*cot(A)²) / tan(A)²
b = sqrt((c² - a²*cot(A)²) / tan(A)²)
 
So I can solve for b in terms of a or a in terms of b...
 
Eh... Is this possible? (OMG I'm such a nerd... wacko.png )

Or maybe something like:

a² + b² = c²
a² = c² - b²
a = sqrt(c² - b²)
tan(A) = a/b
b*tan(A) = a
b*tan(A) = sqrt(c² - b²)

Which eliminates a but I don't know if that b² is recoverable...

Okay, so...

Say A = 36.8699° and c = 10
tan(A) = 3/4

We could say that:
a = 3x
b = 4x

(3x)² + (4x)² = 10²
9x² + 16x² = 100
25x² = 100
x² = 100/25
x² = 4
x = 2

a = 6
b = 8

Which is correct, so this is possible with that method, but now I'm wondering how I can express that as a formula...

Oh, I got it...

Say A = 36.8699° and c = 10
tan(A) = 0.75

a = 0.75b

(0.75b)² + b² = 10²
0.5625b² + b² = 100
1.5625b² = 100
b² = 100/1.5625
b² = 64
b = 8
a = 0.75*8 = 6

Schwing! Achievement unlocked.

Hope someone else enjoys this.

#5Khatharr

Posted 01 February 2013 - 04:35 AM

Calc class is covering pythagoras and trig, which is pretty familiar, so I'm slightly bored which leads to mischief.
 
If I had a right triangle with a known hypotenuse and one angle is it possible to get the lengths of the other two sides without using cos/csc/sin/sec? In other words only using tan/cot and pythagoras?
 
It seems like it should be possible. With pythagoras I can get the sum of a² and b² and with tan/cot I can get the ratio between a and b. I'm fiddling around on paper trying to do it as a formula but it's not simplifying for me.
 
With sides a,b,c and angles A,B,C where the matching letters are opposites (side opposite angle) and c is the hypotenuse, given A and c:
 
a² + b² = c²
(a*b/b)² + (b*a/a)² = c²
(b*a/b)² + (a*b/a)² = c²
(b*tan(A))² + (a*cot(A))² = c²
b²*tan(A)² + a²*cot(A)² = c²
b²*tan(A)² = c² - a²*cot(A)²
b² = (c² - a²*cot(A)²) / tan(A)²
b = sqrt((c² - a²*cot(A)²) / tan(A)²)
 
So I can solve for b in terms of a or a in terms of b...
 
Eh... Is this possible? (OMG I'm such a nerd... wacko.png )

Or maybe something like:

a² + b² = c²
a² = c² - b²
a = sqrt(c² - b²)
tan(A) = a/b
b*tan(A) = a
b*tan(A) = sqrt(c² - b²)

Which eliminates a but I don't know if that b² is recoverable...

Okay, so...

Say A = 36.8699° and c = 10
tan(A) = 3/4

We could say that:
a = 3x
b = 4x

(3x)² + (4x)² = 10²
9x² + 16x² = 100
25x² = 100
x² = 100/25
x² = 4
x = 2

a = 6
b = 8

Which is correct, so this is possible with that method, but now I'm wondering how I can express that as a formula...

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