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Math Help - Find the angle of a triangle ...

  1. #1
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    Find the angle of a triangle ...

    Can you help me finding
    the angle A (see the attachment please.)

    Find the angle of a triangle ...-triangle2.jpg

    I don't know how to start ...
    Last edited by razemsoft21; January 17th 2011 at 03:55 AM.
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  2. #2
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    Well, here's a start!

    A = angle A : so angle C = 100 - A which leaves angle BDC = 40 + A

    a = BC, b = AC, c = AB, d = BD : so AD = a + d

    Insert point E on AD such that DE = d, which leaves AE = a :
    so we now have isosceles triangle BDE, equal sides = d, equal angles = (40 + A) / 2.

    Insert point F on AB, such that EF is parallel to BC :
    so we have triangle AFE, similar to triangle ABC

    I think that should lead you to the solution: give it a shot...
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  3. #3
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    All understood

    but similar triangles treats the sides of
    the triangle, how can I treat the angles ?
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  4. #4
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    Well it seems ABC is isosceles, are you told that?

    If so Angles ABC = 80 and BCA = 80

    Angle BAC = 180 - (ABC - BCA)
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  5. #5
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    Quote Originally Posted by pickslides View Post
    Well it seems ABC is isosceles, are you told that?

    If so Angles ABC = 80 and BCA = 80

    Angle BAC = 180 - (ABC - BCA)
    It should be that, but the question doesn't mention that

    so we have to show that ABC is isosceles,but how ?
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  6. #6
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    Practicing my diagram skills:



    I agree that it is not obvious that AB = AC. Because of this, I think a crucial element of the solution is still missing. In particular, it is not clear how Wilmer's solution uses the fact that AE = a.
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  7. #7
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    I think the condition AD=DC+BCis only true for an isosceles triangle 20-80-80 with a specific base lenght



    bjh
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  8. #8
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    Quote Originally Posted by bjhopper View Post
    I think the condition AD=DC+BCis only true for an isosceles triangle 20-80-80 with a specific base lenght

    bjh
    You are right, but we cann't assume this without proving, can we ?
    and as you know, this is what the question wants us to show.
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  9. #9
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    When two triangles are similar, every length is multiplied by a constant factor. So the relationship AD = BC + BD is preserved.
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  10. #10
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    Yes that was a faulty assumption.I calculated the value of AD,BD,DC and BC for a triangle where AB = 10.BC =3.947 DC =2.578, AD=7.422 BD=3.949
    7.422=3.947 +3.473
    These values were calculated using trig. BC using cos 80/10 =1/2 of BC
    using the angle bisector theorem BC/DC= 10/AD and from this 3.473 /DC = 10/10-DC
    BD is found using the sin rule AC/sin60= BD/sin 80

    Is there a geometry proof? I don't know

    bjh
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  11. #11
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    Quote Originally Posted by razemsoft21 View Post
    Can you help me finding
    the angle A (see the attachment please.)

    Click image for larger version. 

Name:	Triangle2.jpg 
Views:	48 
Size:	34.9 KB 
ID:	20471

    I don't know how to start ...

    (1) Locate Q such that triangles BQD and BCD are congruent.

    (2) Rotate triangle BCD about D until B is at P and C is at Q.

    P, D and C will be collinear only if \alpha=180^o-2\alpha\Rightarrow\ \alpha=60^o

    |\angle\ DQA|=\alpha+40^o\Rightarrow\ |\angle\ QAP|=180^o-\left[180^o-2\alpha+\alpha+40^o\right]=\alpha-40^o

    Therefore A, Q and B are collinear and triangle APQ is isosceles only if \alpha=60^o

    \Rightarrow\ |\angle\ QPA|=140^o\Rightarrow\ |\angle\ QAP|=20^o
    Attached Thumbnails Attached Thumbnails Find the angle of a triangle ...-razemsoft.jpg  
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  12. #12
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    Quote Originally Posted by bjhopper View Post
    Yes that was a faulty assumption.I calculated the value of AD,BD,DC and BC for a triangle where AB = 10.BC =3.947 DC =2.578, AD=7.422 BD=3.949
    7.422=3.947 +3.473
    These values were calculated using trig. BC using cos 80/10 =1/2 of BC
    using the angle bisector theorem BC/DC= 10/AD and from this 3.473 /DC = 10/10-DC
    BD is found using the sin rule AC/sin60= BD/sin 80

    Is there a geometry proof? I don't know

    bjh
    again you are right

    but you cann't generalized this special case using numbers
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  13. #13
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    Something I don't understand...
    Quote Originally Posted by Archie Meade View Post
    (1) Locate Q such that triangles BQD and BCD are congruent.

    (2) Rotate triangle BCD about D until B is at P and C is at Q.
    At this point we don't know if it is possible for the rotated image of C to lie on AB and the image of B to lie on AC. Since above you defined Q but not P, I assume that we rotate BCD until C coincides with Q. The image of B, i.e., P, does not have to be on AC.

    P, D and C will be collinear only if \alpha=180^o-2\alpha\Rightarrow\ \alpha=60^o
    Right, so if we knew that P, D and C are collinear, i.e., that P lies on AC, then the proof would be finished.

    |\angle\ DQA|=\alpha+40^o\Rightarrow\ |\angle\ QAP|=180^o-\left[180^o-2\alpha+\alpha+40^o\right]=\alpha-40^o
    This is I am not sure about. I see how we find the angle QAD in the triangle QAD by subtracting angles Q and D from 180. However, it is not clear that \angle QAD=\angle QAP since P may potentially not lie on AD.

    Therefore A, Q and B are collinear and triangle APQ is isosceles only if \alpha=60^o
    That A, Q and B are collinear is known, but why AP = PQ is not clear (for the same reason as above).

    Quote Originally Posted by bjhopper
    I calculated the value of AD,BD,DC and BC for a triangle where AB = 10.BC =3.947 DC =2.578, AD=7.422 BD=3.949
    7.422=3.947 +3.473
    These values were calculated using trig. BC using cos 80/10 =1/2 of BC
    using the angle bisector theorem BC/DC= 10/AD and from this 3.473 /DC = 10/10-DC
    BD is found using the sin rule AC/sin60= BD/sin 80
    I am not sure I understand this either. It seems that you assumed that \angle BAC=20^\circ and that the triangle ABC is isosceles to find its sides, whereas the problem was to prove that \angle A=20. In particular:
    cos 80/10 =1/2 of BC
    (You must mean (cos 80) * 10 = 1/2 BC.) This assumes that the perpendicular from A divides BC is in half, which is not obvious. Also:
    from this 3.473 /DC = 10/10-DC
    By assuming that AD = 10 - DC, you assume that AC = AB.

    I also came up with a trigonometry solution. Let \angle BAC=\alpha. By the law of sines for BAD, \displaystyle\frac{BD}{\sin\alpha}=\frac{AD}{\sin 40}=\frac{BD+BC}{\sin 40}. Also, for BDC, \displaystyle\frac{BD}{\sin(100-\alpha)}=\frac{BC}{\sin(40+\alpha)}. Now, we may give BD any numeric value; then BC and \alpha will be determined uniquely because the whole picture can be scaled arbitrarily preserving \alpha. So, let BD = 1 and denote BC by a. From the first equation, a=(\sin40)/(\sin\alpha)-1; from the second one, a=(\sin(40+\alpha))/(\sin(100-\alpha)). Therefore,

    \displaystyle f(\alpha):=\frac{\sin40}{\sin\alpha}-\frac{\sin(40+\alpha)}{\sin(100-\alpha)}=1

    It is easy to see that on (0, 100), the function f(\alpha) is strictly decreasing: the denominator of the first fraction and the nominator of the second one increase while the denominator of the second fraction decreases. Therefore, there is at most one solution. With a bit of trig, it is possible to show that f(20) = 1.
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    Quote Originally Posted by emakarov View Post
    Something I don't understand...
    At this point we don't know if it is possible for the rotated image of C to lie on AB

    This we do know as |QD|=|DC| since triangles DBC and DQB have x, y and 40 degrees as shown.
    Then we check the condition required for P to be on AC.
    This needs the angle "alpha" to be 60 degrees.

    Then we examine the conditions for collinearity of A, Q, B.



    and the image of B to lie on AC. Since above you defined Q but not P, I assume that we rotate BCD until C coincides with Q. The image of B, i.e., P, does not have to be on AC.

    Right, so if we knew that P, D and C are collinear, i.e., that P lies on AC, then the proof would be finished.
    .
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  15. #15
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    Sorry, I spent a lot of time on this, and I am still confused. Could you write more precisely how and in which order you define various objects and what you know about them at that time, not what you would like to prove eventually?

    At this point we don't know if it is possible for the rotated image of C to lie on AB

    This we do know as |QD|=|DC| since triangles DBC and DQB have x, y and 40 degrees as shown.
    Then we check the condition required for P to be on AC.
    We can have for free that the image of C is Q (so lies on AB) or that the image of B lies on AC. I thought you chose the first thing, making Q the image of C by definition. That's fine. But the rest of the proof uses the fact that the image of B (I presume, it is P by definition) is on AC, and I don't see it proved.
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