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    So I was revising for my End Of Year exam which will be a mock of the REAL new GCSE and this question came up.
    Anyone know , step by step, how to solve this?

    Thanks
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    (Original post by NothingButWaleed)
    So I was revising for my End Of Year exam which will be a mock of the REAL new GCSE and this question came up.
    Anyone know , step by step, how to solve this?
    can't see the question
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    (Original post by fatima1998)
    can't see the question
    Its there now!
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    Draw line from A to C and one from A horizontally across to the line DC. You then get a right angled triangle and use the angle along with the ratio to get sides etc.
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    Work out the other side using the ratio.
    Draw the angle of elevation and then draw a right angled triangle connecting the two lines together. You will realise the angle of elevation is the same as one of the angles - corresponding angles ?

    Use SOH CAH to then work it out.
    Well, in all honesty you can only use tan or TOA haha


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    (Original post by Chittesh14)
    Work out the other side using the ratio.
    Draw the angle of elevation and then draw a right angled triangle connecting the two lines together. You will realise the angle of elevation is the same as one of the angles - corresponding angles ?

    Use SOH CAH to then work it out.
    Well, in all honesty you can only use tan or TOA haha


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    I just tried it but no luck. I do something wrong but I dont know what, can you do the working out and show me?
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    CD = 1.5 x 1.7 = 2.55

    Let E be the point where a perpendicular line to AB, at A intersects CD

    CE = CD - AB = 2.55 - 1.7 = 0.85

    To work out EA use trig:

    Tan(angle) = opp/adj
    Tan(52) = (0.85)/EA
    EA = (0.85)/tan(52)
    EA = 0.6640927... = 0.664 (to 3sf)

    EA = BD
    Therefore BD = 0.664 (to 3sf)

    Posted from TSR Mobile
 
 
 
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