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.
