Math problem

Nabs

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?

xterm

CIB = 180 - 45 = 135
ICB = 180 - 135 - 30 = 15
ACB = ICB * 2 = 30 (Since I is the midpoint of AB)

CAB = 180 - ABC - ACB = 180 - 30 - 30 = 120

That may be right >.>

Nabs

ACE is not equal to ICB*2, because that is any triangle, so the median does not bisect the vertex in half.

xterm

I'll let you read the given again :-)

Tarek

The data are all messed up!

Nabs

@Tarek, messed up how?

Nabs

YIPES, sorry, the info is wrong. Let me update please. I am so sorry.

Tarek

They said CAB IS 30 degree then ask you to find CAB

Nabs

Here's the right info:

Tarek

I can do it by calculus it gives 120 degrees

Nabs

How?

Tarek

Express CAB in the terms of other angles.. Une mediane peut etre considerer comme bissectrice?

Nabs

@Tarek, a median only bisects a vertex in an isosceles or equilateral triangle. Doesn't work here.

Hybrid

AIB = 180
CIB = 180 - 45 = 135
ICB = 180 - 30 - 135 = 15
ICA = ICB = 15
IAC = 180 - 15 - 45 = 120

Not sure about the last two, but it sounds correct :P

Nabs

@Paladin, ICA is not equal to ICB. CI is a median, not a bisector.

Metalloy

A=105
hint: use the sin law on all three triangles and try to find relations

arithma

ICA = ICB is not a correct assumption.

Fixed a couple errors:
- Was moving the line to the right by adding to x (ie (x+1)) instead of subtracting from it
- Was drawing lines with a slope of Sin[Angle] rather than Tan[Angle]. Stupid.

Metalloy got it right from the beginning!

mesa177

Hmmm, that's not what I got:

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(ACI)/sin(CAI)
=> IB*sin(CBI)/sin(ICB) = AI*sin(ACI)/sin(CAI) (eq1)

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

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

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

sin(a-b) = sin(a)cos(b) - cos(a)sin(b)
=> sin(CAB) = [sin(135)cos(CAB) - cos(135)sin(CAB)]*sin(15)/sin(30)
divide by cos(CAB)
=> tan(CAB) = [sin(135) - cos(135)tan(CAB)]*sin(15)/sin(30)
=> 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

Am I missing something?

Metalloy

mesa177 wrote => tan(CAB) = [sin(135) - cos(135)tan(CAB)]*sin(15)/sin(30)
=> 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

Am I missing something?

I didnt get the transition made from the first to the second equation from those I quoted, i dont have a paper in frobt of me to verify if it is correct but maybe there's a mistake in here?

rtp

hmmm 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

xterm gave the simple straight solution.
120 moving on already