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    Hey, really stuck on this q, any help or clues to get started would be appreciated.

    "In the diagram AB=6cm is the diamater of the circle and BT is the tangent to the circle at B. The chord AC is extened to meet this tangent at D and <DAB= theta
    a) Show that CD = 6(sec theta - cos theta)"

    If anyone has the edexcel C3 book the diagram is on p/82 and q.8
    Thnaks
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    (Original post by jimber)
    Hey, really stuck on this q, any help or clues to get started would be appreciated.

    "In the diagram AB=6cm is the diamater of the circle and BT is the tangent to the circle at B. The chord AC is extened to meet this tangent at D and <DAB= theta
    a) Show that CD = 6(sec theta - cos theta)"

    If anyone has the edexcel C3 book the diagram is on p/82 and q.8
    Thnaks
    ok the first thing to do is connect C to B so that you have a triangle ACB where angle acb =90 since angle in a semicircle is 90.And triangle ABD is also 90 , diameter is perpendicular to tangent. Now using trig find AD for triangle ABC , which is 6sectheta , similary find AC in triangle ABD which is 6costheta.
    so CD= 6(sectheta -costheta)
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    I got CD = 6* SinTheta*TanTheta by working out BD first

    Presumably that is the same thing.
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    (Original post by rbnphlp)
    ok the first thing to do is connect C to B so that you have a triangle ACB where angle acb =90 since angle in a semicircle is 90.And triangle ABD is also 90 , diameter is perpendicular to tangent. Now using trig find AD for triangle ABC , which is 6sectheta , similary find AC in triangle ABD which is 6costheta.
    so CD= 6(sectheta -costheta)
    Hey, thanks for the reply. Big help! I'll rep you in a mo
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    (Original post by jimber)
    Hey, thanks for the reply. Big help! I'll rep you in a mo
    thnx for the rep
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    (Original post by stevencarrwork)
    I got CD = 6* SinTheta*TanTheta by working out BD first

    Presumably that is the same thing.
    Hey thanks, how did you work out BD? Just realised I have to use this method to answer the next part to the question lol
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    (Original post by jimber)
    Hey thanks, how did you work out BD? Just realised I have to use this method to answer the next part to the question lol
    DB and AB are at right angles. So BD/AB = tan Theta
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    (Original post by stevencarrwork)
    DB and AB are at right angles. So BD/AB = tan Theta
    Thanks, the next part of the question told me CD was 16cm and asked if i cud calculate AC, but i thought i would need the length of AD to do this?
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    You have a formula for CD, so presumably you solve it for theta.

    What sort of triangle is ACB?
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    (Original post by stevencarrwork)
    You have a formula for CD, so presumably you solve it for theta.

    What sort of triangle is ACB?
    I've got 6(sec theta - cos theta) = 16 but i'm not sure how to solve this, tried by multiplying by cos and replacing the cos's with x's and solving as a quadratic but it failed lol.

    umm ACB is just a normal triangle, no right-angles but i get the feeling angle >CAB = >ABT?
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    If AB is a diameter, what angle is at C?
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    (Original post by stevencarrwork)
    If AB is a diameter, what angle is at C?
    DOH! i'm officialy retarded lol sorry about the late reply, but don't i still need to know one more thing before i can work out AC?
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    (Original post by jimber)
    I've got 6(sec theta - cos theta) = 16 but i'm not sure how to solve this, tried by multiplying by cos and replacing the cos's with x's and solving as a quadratic but it failed lol.
    It can't fail. The quadratic can be solved if you do what you did.
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    (Original post by stevencarrwork)
    It can't fail. The quadratic can be solved if you do what you did.
    raaa i must've made a mistake, for my quadratic i got:
    x^2 - (16x/6) - 1 = 0
    I tried to use the quadratic equation but it gave me answers >1
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    You must have a minus sign wrong.
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    (Original post by stevencarrwork)
    You must have a minus sign wrong.
    Yeah I did, thanks for your help, i'll rep you in a mo
 
 
 
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