Jump to content

View more

Image of the Day

Inventory ! Va falloir trouver une autre couleur pour le cadre D: #AzTroScreenshot #screenshotsaturday https://t.co/PvxhGL7cOH
IOTD | Top Screenshots

The latest, straight to your Inbox.

Subscribe to GameDev.net Direct to receive the latest updates and exclusive content.


Sign up now

Abusing tan/cot for fun and profit...

4: Adsense

Old topic!

Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.


  • You cannot reply to this topic
2 replies to this topic

#1 Khatharr   Members   

8639
Like
0Likes
Like

Posted 01 February 2013 - 03:43 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.


Edited by Khatharr, 01 February 2013 - 06:18 AM.

void hurrrrrrrr() {__asm sub [ebp+4],5;}

There are ten kinds of people in this world: those who understand binary and those who don't.

#2 Álvaro   Members   

20922
Like
0Likes
Like

Posted 01 February 2013 - 06:54 AM

If you know the sum of the squares of a and b and their ratio, it's easy to solve the system
a^2+b^2 = X
a/b = Y

a = b*Y
(b*Y)^2+b^2 = X
(1+Y^2)*b^2 = X
b = sqrt(X/(1+Y^2))
a = Y*sqrt(X/(1+Y^2))

#3 Khatharr   Members   

8639
Like
0Likes
Like

Posted 01 February 2013 - 07:01 AM

It was easy to do when I applied it to a real triangle, but before that I was trying to do it all with formulas and it wasn't working. Once I did it on a triangle with known sides and angles I was able to express the process.

 

I'm hoping that there will be another batch of problems posted before the end of the unit so I can use this method rather than sin or cos and see if my professor flips out. :D


Edited by Khatharr, 01 February 2013 - 07:02 AM.

void hurrrrrrrr() {__asm sub [ebp+4],5;}

There are ten kinds of people in this world: those who understand binary and those who don't.




Old topic!

Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.