specialist question

[Biceps]

Tan(a)=1/12 Tan(B)=2/5 Tan(C)=1/3

Find a+B+C

[d3stiny]

tan(a+b) = tan(a)+tan(b) / (1 - tan(a)tan(b)) = (1/12 + 2/5) / (1-1/60) = 1/2
tan(a+b+c) = tan(a+b) + tan(c) / (1 - tan(a+b)tan(c)) = 1/2 + 1/3 / (1 - 1/6) = 1
so a+b+c = tan^-1 (1) = pi/4

[pi]

This was the method i used, much more tedious than above, but i typed it out so meh





Ok, assume

Therefore

Firstly, tackle , let that sum be













To solve for













Hence,

[Special At Specialist]

tan(a) = 1/12, tan(b) = 2/5, tan(c) = 1/3
First we should recognise that 0 < a < c < b < pi/4
The reason is because tan(x) is an increasing function for 0 < x < pi/2 and tan(0) = 0 and tan(pi/4) = 1

tan(a + b) = (tan(a) + tan(b)) / (1 - tan(a)tan(b))
tan(a + b) = (1/12 + 2/5) / (1 - (1/12)*(2/5))
Multiply numerator and denominator by 60:
tan(a + b) = (5 + 24) / (60 - 2)
tan(a + b) = 29 / 58
tan(a + b) = 1/2
a + b = arctan(1/2), since 0 < (a + b) < pi/2

tan(b + c) = (tan(b) + tan(c)) / (1 - tan(b)tan(c))
tan(b + c) = (2/5 + 1/3) / (1 - (2/5)(1/3))
Multiply numerator and denominator by 15:
tan(b + c) = (6 + 5) / (15 - 2)
tan(b + c) = 11 / 13
b + c = arctan(11/13), since 0 < (b + c) < pi/2

tan(a + c) = (tan(a) + tan(c)) / (1 - tan(a)tan(c))
tan(a + c) = (1/12 + 1/3) / (1 - (1/12)(1/3))
Multiply numerator and denominator by 36:
tan(a + c) = (3 + 12) / (36 - 1)
tan(a + c) = 15 / 35
tan(a + c) = 3/7
a + c = arctan(3/7), since 0 < (a + c) < pi/2

Now just add all of the equations together:
(a + b) + (b + c) + (a + c) = arctan(1/2) + arctan(11/13) + arctan(3/7)
2a + 2b + 2c = arctan(1/2) + arctan(11/13) + arctan(3/7)
a + b + c = (1/2)*(arctan(1/2) + arctan(11/13) + arctan(3/7))

That is your answer, though you might be able to simplify things a bit:
Let u = arctan(1/2) + arctan(11/13)
tan(u) = tan(arctan(1/2) + arctan(11/13))
Using the rule that tan(arctan(x)) = x and tan(a + b) = (tan(a) + tan(b)) / (1 - tan(a)tan(b)):
tan(u) = (1/2 + 11/13) / (1 - (1/2)(11/13))
Multiply numerator and denominator by 26:
tan(u) = (13 + 22) / (26 - 11)
tan(u) = 35 / 15
tan(u) = 7 / 3
u = arctan(7/3), since 0 < u < pi/2

Therefore:
a + b + c = (1/2)*(arctan(7/3) + arctan(3/7))

Let v = arctan(7/3) + arctan(3/7)
tan(v) = tan(arctan(7/3) + arctan(3/7))
tan(v) = (7/3 + 3/7) / (1 - (7/3)(3/7))
Multiply numerator and denominator by 21:
tan(v) = (49 + 9) / (21 - 21)
tan(v) = 58 / 0
tan(v) = undefined
Therefore v = pi/2

a + b + c = (1/2)*(pi/2)
a + b + c = pi/4

[Jenny_2108]

This question is so much fun so I wanna contribute one more method This formula is out of course though



Sub in, we get pi/4

[Biceps]

Thanks for responses guys I really appreciate it.