Math problem

Nabs · 2013-03-14T20:49:42+00:00

OK, this is a bit embarrassing for me, but in my defense, I don't have my full memory on me. Angle A, how much is it?

Nabs

@Mesa177, the answer is not 30.

Metalloy

Nabs wrote@Mesa177, the answer is not 30.

do you have the right answer?

Ayman

rtp wrotehmmm am not sure why this is going on with tan,sin and cos. What the hell.
There should be a theory that states that ACI = ICB, since I is the midpoint thus it puts the angle C in half or something

Wrong. This can be true only if the triangle was equilateral, the segment CI would then form a bisector to the angle as it intersects AB in the midpoint and forms a 90 angle there on intersection. So its more of a special case.

rtp

@Ayman, ok man. So it is complicated...

Never the less, am going to be a smart pants and say it can't be solved

mesa177

Nabs wrote@Mesa177, the answer is not 30.

Yeah, I know, after drawing the triangle with a compass and protractor, angle CAB is obtuse and equal 105 degrees. My main question is why on earth did I reach 30?

@Metalloy: The detailed steps are:

=> tan(CAB) = [sin(135) - cos(135)tan(CAB)]*sin(15)/sin(30)
=> tan(CAB)*sin(30) = [sin(135) - cos(135)tan(CAB)]*sin(15)
=> tan(CAB)*sin(30) = sin(135)sin(15) - cos(135)sin(15)*tan(CAB)
=> tan(CAB)*sin(30) + cos(135)sin(15)*tan(CAB) = sin(135)sin(15)
=> tan(CAB) [sin(30) + cos(135)sin(15)] = sin(135)sin(15)
=> tan(CAB) = sin(135)sin(15)/[sin(30) + cos(135)sin(15)]
=> CAB = arctan{sin(135)sin(15)/[sin(30) + cos(135)sin(15)]} = 30

maulader

mesa177 wrote and IC/sin(CAI) = AI/sin(ACI) = CA/sin(AIC)
=> IC = AI*sin(ACI)/sin(CAI)

correct me if i'm wrong but i think u swapped them :P

Metalloy

maulader wrote
mesa177 wrote and IC/sin(CAI) = AI/sin(ACI) = CA/sin(AIC)
=> IC = AI*sin(ACI)/sin(CAI)
correct me if i'm wrong but i think u swapped them :P

correct

mesa177

Yes I did, stupid mistake indeed...

So here's the final solution:

CIB = 180 - CIA = 180 - 45 = 135
ICB = 180 - (CIB + CBI) = 180 - 165 = 15

sine rule states that
CB/sin(CIB) = IB/sin(ICB) = IC/sin(CBI)
=> IC = IB*sin(CBI)/sin(ICB)
and IC/sin(CAI) = AI/sin(ACI) = CA/sin(AIC)
=> IC = AI*sin(CAI)/sin(ACI)
=> IB*sin(CBI)/sin(ICB) = AI*sin(CAI)/sin(ACI) (eq1)

but AI = IB
=> sin(CBI)/sin(ICB) = sin(CAI)/sin(ACI)
since CAI = CAB (A, I and B are collinear)
=> sin(CBI)/sin(ICB) = sin(CAB)/sin(ACI)
=> sin(CAB) = sin(ACI)*sin(CBI)/sin(ICB)
=> sin(CAB) = sin(ACI)*sin(30)/sin(15)
but ACI = ACB - ICB = ACB - 15
=> sin(CAB) = sin(ACB - 15)*sin(30)/sin(15) (eq2)

also ACB = 180 - (CAB + ABC) = 180 - CAB - 30
=> ACB = 150 - CAB (eq3)

from eq1 and eq2:
sin(CAB) = sin(150 - CAB - 15)*sin(30)/sin(15)
=> sin(CAB) = sin(135 - CAB)*sin(30)/sin(15) (eq3)

sin(a-b) = sin(a)cos(b) - cos(a)sin(b)
=> sin(CAB) = [sin(135)cos(CAB) - cos(135)sin(CAB)]*sin(30)/sin(15)
divide by cos(CAB)
=> tan(CAB) = [sin(135) - cos(135)tan(CAB)]*sin(30)/sin(15)
=> tan(CAB) [sin(15) + cos(135)sin(30)] = sin(135)sin(30)
=> tan(CAB) = sin(135)sin(30)/[sin(15) + cos(135)sin(30)]
=> CAB = arctan{sin(135)sin(30)/[sin(15) + cos(135)sin(30)]}
=> CAB = -75 or 180 - (-75) = 105 (arctan(a) = b or 180 - b)
but in traingle there are no negative angles
=> CAB = 105

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