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GCSE Triangles question, need help please! watch

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    I'm asking this question on behalf of my little bro since he doesn't have an account in here:

    Name:  image.jpg
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    Hope the picture is ok...

    Any hints for my bro would be great. Thanks. (Ignore my little bro's workings)
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    bump!
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    (Original post by jordanwu)
    I'm asking this question on behalf of my little bro since he doesn't have an account in here:

    Name:  image.jpg
Views: 174
Size:  354.6 KB

    Hope the picture is ok...

    Any hints for my bro would be great. Thanks. (Ignore my little bro's workings)
    Has you little bro done SOH CAH TOA?
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    (Original post by maggiehodgson)
    Has you little bro done SOH CAH TOA?
    There are no given sides, how do you use trig for it?
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    (Original post by jordanwu)
    There are no given sides, how do you use trig for it?
    Well the little triangle is a right angled isos. So you could sa that AB is the same length as the base. Let's say 1 unit. You are told that the AB = BC so AC must be 2 units with the same 1 unit base.

    If it's not done with trig I've no idea how to do it any other way.
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    (Original post by maggiehodgson)
    Well the little triangle is a right angled isos. So you could sa that AB is the same length as the base. Let's say 1 unit. You are told that the AB = BC so AC must be 2 units with the same 1 unit base.

    If it's not done with trig I've no idea how to do it any other way.
    So what would the value of the lengths actually be? I'm pretty sure you use trig for it.
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    (Original post by jordanwu)
    So what would the value of the lengths actually be? I'm pretty sure you use trig for it.
    If the base was 1 unit the the perpendicular of the big triangle must be 2.
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    (Original post by maggiehodgson)
    If the base was 1 unit the the perpendicular of the big triangle must be 2.
    No need to use trig.

    Its just using angles in triangles, the fact that there are 360 degrees in a quadrilateral CXBY and the fact that angle CXB=CYB
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    (Original post by aoxa)
    No need to use trig.

    Its just using angles in triangles, the fact that there are 360 degrees in a quadrilateral CXBY and the fact that angle CXB=CYB
    Does it say that CX =CY?
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    (Original post by maggiehodgson)
    Does it say that CX =CY?
    No, but given the symmetry in the picture, and that this is a gcse question, where they will not give trig when there are so many unknowns, I would say it can be assumed.
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    (Original post by aoxa)
    No, but given the symmetry in the picture, and that this is a gcse question, where they will not give trig when there are so many unknowns, I would say it can be assumed.
    The question does say "Not to scale". I would say that you shouldn't go by the picture only the description of the diagram. But, I might be wrong.
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    (Original post by maggiehodgson)
    The question does say "Not to scale". I would say that you shouldn't go by the picture only the description of the diagram. But, I might be wrong.
    Not to scale, as in don't get your protractor out to measure the angle. If the diagram wasn't supposed to be symmetrical, then the diagram would have made that way more obvious.
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    (Original post by aoxa)
    No, but given the symmetry in the picture, and that this is a gcse question, where they will not give trig when there are so many unknowns, I would say it can be assumed.
    What's angle XCY?
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    (Original post by maggiehodgson)
    What's angle XCY?
    45 I think. I haven't done geometrical shapes since gcse, but I think I recall that angle XBY is double angle XCY
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    (Original post by aoxa)
    45 I think. I haven't done geometrical shapes since gcse, but I think I recall that angle XBY is double angle XCY
    I think that must be when it's in a circle and B is the centre.

    If you were right, and the diagram is symmetrical, then cxb would be 22.5. That would mean that XB = CB but as CB = AB it would mean that the hypotenuse of the little triangle was the same as one of its other sides.
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    (Original post by maggiehodgson)
    I think that must be when it's in a circle and B is the centre.

    If you were right, and the diagram is symmetrical, then cxb would be 22.5. That would mean that XB = CB but as CB = AB it would mean that the hypotenuse of the little triangle was the same as one of its other sides.
    Hmm but the question doesn't say that it's a circle..
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    (Original post by aoxa)
    45 I think. I haven't done geometrical shapes since gcse, but I think I recall that angle XBY is double angle XCY
    Not sure if circle theorems can be used in this question since the triangles aren't in a circle... but if they were then x would be 67.5 degrees
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    (Original post by jordanwu)
    Not sure if circle theorems can be used in this question since the triangles aren't in a circle... but if they were then x would be 67.5 degrees
    Seems correct but I am very tired so arithmetic mistakes are possible, try talk your bro through this, don't give him the answer straight away

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    (Original post by jordanwu)
    I'm asking this question on behalf of my little bro since he doesn't have an account in here:

    Name:  image.jpg
Views: 174
Size:  354.6 KB

    Hope the picture is ok...

    Any hints for my bro would be great. Thanks. (Ignore my little bro's workings)
    I'm not sure if this has been said already:

    If you let XA = m.

    Then AB = m since ABX is isosceles and AC = 2m.


    So you can use trig on triangle ACX:

    \displaystyle \tan (CXA) = \frac{2m}{m} = ...
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    (Original post by notnek)
    I'm not sure if this has been said already:

    If you let XA = m.

    Then AB = m since ABX is isosceles and AC = 2m.


    So you can use trig on triangle ACX:

    \displaystyle \tan (CXA) = \frac{2m}{m} = ...
    Thanks for the confirmation of my solution in a much earlier post. I was having a hard time convincing one poster that it wasn't a geometry solution.
 
 
 
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