Hey there! Sign in to join this conversationNew here? Join for free
    Offline

    4
    ReputationRep:
    If only there was a circle around the triangle, ez 150/2 :laugh:
    Offline

    18
    ReputationRep:
    (Original post by morgan8002)
    I'm pretty confused. My last line of working was \cos \frac{\theta}{2} = 0, which I wrote down the wrong solution \theta = 90^o. There was something else that could have been 0 instead which I'm working on now, but I don't see any reason why I would accidentally come up with the right answer.
    :dontknow: if it's stupid but works, then it ain't stupid!
    Spoiler:
    Show
    *applies to all fields of life except mathematics :getmecoat:
    Offline

    12
    ReputationRep:
    (Original post by morgan8002)
    I'm pretty confused. My last line of working was \cos \frac{\theta}{2} = 0, which I wrote down the wrong solution \theta = 90^o. There was something else that could have been 0 instead which I'm working on now, but I don't see any reason why I would accidentally come up with the right answer.
    what did you get as the lengths of triangle?
    Offline

    14
    ReputationRep:
    Let φ=CBA=BCA
    AD=1 and we take a parallel line to BC from D which intersects CA at E
    DE= x
    cosφ = x/2
    sinφ = (x+sqrt(6))tan15/sqrt((sqrt(6)-x)^2+(x+sqrt(6))^2*tan^2(15))
    sin^2φ+cos^2φ=1 which gives x thus φ thus θ
    Offline

    18
    ReputationRep:
    (Original post by Vesniep)
    Let φ=CBA=BCA
    AD=1 and we take a parallel line to BC from D which intersects CA at E
    DE= x
    cosφ = x/2
    sinφ = (x+sqrt(6))tan15/sqrt((sqrt(6)-x)^2+(x+sqrt(6))^2*tan^2(15))
    sin^2φ+cos^2φ=1 which gives x thus φ thus θ
    :coma:
    Offline

    14
    ReputationRep:
    (Original post by Student403)
    :coma:
    I know
    Offline

    18
    ReputationRep:
    (Original post by Vesniep)
    I know
    Very humble :rofl:
    • Community Assistant
    • Study Helper
    • Thread Starter
    Offline

    20
    ReputationRep:
    Community Assistant
    Study Helper
    (Original post by Vesniep)
    Let φ=CBA=BCA
    AD=1 and we take a parallel line to BC from D which intersects CA at E
    DE= x
    cosφ = x/2
    sinφ = (x+sqrt(6))tan15/sqrt((sqrt(6)-x)^2+(x+sqrt(6))^2*tan^2(15))
    sin^2φ+cos^2φ=1 which gives x thus φ thus θ
    I think I did this and ended up with a quartic. Is that what you got?
    Offline

    14
    ReputationRep:
    (Original post by notnek)
    I think I did this and ended up with a quartic. Is that what you got?
    yes it is a quartic , then you ask help from geogebra and the work is done .
    • Community Assistant
    • Study Helper
    • Thread Starter
    Offline

    20
    ReputationRep:
    Community Assistant
    Study Helper
    So it looks this can be solved by getting help to solve a quartic, which is what I thought.

    The challenge is to find a nicer method that gives the answer directly. It could be that there isn't one.
    Offline

    14
    ReputationRep:
    (Original post by notnek)
    So it looks this can be solved by getting help to solve a quartic, which is what I thought.

    The challenge is to find a nicer method that gives the answer directly. It could be that there isn't one.
    Did you find a solution ?
    I found two roots for that awful equation .
    0<φ < 90 so I eliminated the negative root and the positive was x=1.5
    cosφ=0.75 thus θ=7.18.
    Perhaps I did something wrong of course.
    The thing is the roots are only two not four so we either have two imaginary roots or they are both double roots which means that the quartic was not that awful.
    • Community Assistant
    • Study Helper
    • Thread Starter
    Offline

    20
    ReputationRep:
    Community Assistant
    Study Helper
    (Original post by Student403)
    Wha?

    I just checked using the sine and cosine rules and 90 was fine
    90 works? Now I'm confused.

    I'll have another look later when I have time.
    Offline

    18
    ReputationRep:
    (Original post by notnek)
    90 works? Now I'm confused.

    I'll have another look later when I have time.
    Unless I'm being stupid, which isn't an unreasonable possibility
    Offline

    19
    ReputationRep:
    absolutely vile
    Offline

    13
    ReputationRep:
    I had a go and got:
    Spoiler:
    Show
    Name:  Possible Z-shape triangle solution.png
Views: 48
Size:  15.3 KB
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Brussels sprouts
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.