November 22nd, 2017, 01:54 PM  #1 
Newbie Joined: Nov 2017 From: Serbia Posts: 6 Thanks: 0  Is this possible to solve?Geometry/Triangle
So my teacher gave this math problem to us so we can solve it; he said that it is very hard even for him to do it.The person who solves it till tomorrow will get 2x A (double A's, double 5). I can't do it; I tried everything, but it just seems impossible. I'm 17 y.o., third grade in high school. I hope someone can help me with this. >We need to find those two angles that are marked on the picture; everything we know is marked and numbers on the picture are angles in degrees. Can you solve this and tell me how you get those answers? I need this ASAP. Thanks in advance! Last edited by skipjack; November 22nd, 2017 at 05:12 PM. 
November 22nd, 2017, 05:08 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,065 Thanks: 1621 
Check your diagram with the teacher. If the diagram is correct, the problem is unfairly difficult. See this discussion. HardTriangle.jpg In the diagram above, the angles are 32° and 18°, but the discussion above indicates these values are only approximately correct. 
November 23rd, 2017, 01:10 AM  #3  
Newbie Joined: Nov 2017 From: Serbia Posts: 6 Thanks: 0  Quote:
 
November 23rd, 2017, 02:36 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 19,065 Thanks: 1621 
There isn't an easier way for this problem. However, I doubt that your teacher intended to set such a hard problem.

November 23rd, 2017, 11:09 AM  #5 
Newbie Joined: Nov 2017 From: Serbia Posts: 6 Thanks: 0 
Oh, I guess that's the case then. Can someone help me to understand how to solve this problem step by step? I know I have the answer, but it's to rush up for me. If someone can help me with step by step howtodo, I will be very thankful.
Last edited by skipjack; November 23rd, 2017 at 05:36 PM. 
November 24th, 2017, 03:38 PM  #6 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,621 Thanks: 845 
How in hell is a student suppose to learn from those convoluted equations shown at "Cut the Knot"?! Actually easy "going the long way". I'll use F instead of your "O"; hate using O. Let AB = 1 TriangleABF: Calculate AF (Sine Law) AF = SIN(20) / SIN(130) = ~.446476 TriangleABF: Calculate BF (Sine Law) BF = (SIN(30) / SIN(130) = ~.652270 TriangleAEF: Calculate EF (Sine Law) (notice that triangle AEF is isosceles) EF = AF * SIN(50) / SIN(80) = ~.347296 TriangleBDF: Calculate DF (Sine Law) DF = BF * SIN(60) / SIN(70) = ~.601534 TriangleDEF: Calculate DE (Cosine Law) DE = SQRT[DF^2 + EF^2  2*DF*EF*COS(130) = ~.866666 TriangleDEF: Calculate angle DEF (Sine Law) angleDEF = ASIN[DF*SIN(130)/DE] = ~32.122012 angle EDF = 180  130  angle DEF = ~17.877987 Methinks that triangle is perfect in order for students to learn about Sin/Cos laws. Last edited by skipjack; November 24th, 2017 at 08:18 PM. 
November 24th, 2017, 03:59 PM  #7 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 1,962 Thanks: 132 Math Focus: Trigonometry  
November 24th, 2017, 04:12 PM  #8 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,621 Thanks: 845 
Looks too much like a zero. Similarly hate using I : looks like one. 
November 24th, 2017, 05:03 PM  #9 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 602 Thanks: 82  That shows that the diagram isn't to scale. In a diagram drawn to scale with the congruent angles being smaller than the other one (50, 50, and 80 in this case), the shortest side is in between the two congruent angles. EO (what you called EF) looks like the shortest side, but AO (what you called AF) is in between the congruent angles.

November 24th, 2017, 06:06 PM  #10 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,621 Thanks: 845 
Well, with top angle = 20, the triangle is a long skinny minny; so the whole shebang is way outta scale!! 

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