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Liv1234

  • one year ago

SOMEONE PLEASE HELP ME WITH A FEW QUESTIONS

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  1. Liv1234
    • one year ago
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  2. kc_kennylau
    • one year ago
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    The tangent-secant theorem: \(CD^2 = CB \times CA\)

  3. Liv1234
    • one year ago
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    @kc_kennylau Can you help me through the problem?

  4. Liv1234
    • one year ago
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    @kc_kennylau

  5. Liv1234
    • one year ago
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    Or anyone that can help please.

  6. Godlovesme
    • one year ago
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    i'll help! :)

  7. Godlovesme
    • one year ago
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    let's use the formula @kc_kennylau provided :3 \[CD^2 = BC+ AC\]

  8. Godlovesme
    • one year ago
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    @Liv1234 u there?

  9. Liv1234
    • one year ago
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    Sorry, yes I'm here.

  10. kc_kennylau
    • one year ago
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    @Godlovesme that was times

  11. Liv1234
    • one year ago
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    Thank you for helping by the way.

  12. Godlovesme
    • one year ago
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    oh sorry :/

  13. kc_kennylau
    • one year ago
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    it's alright

  14. Liv1234
    • one year ago
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    So, how would I use the formula to find out the answer?

  15. Liv1234
    • one year ago
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    @Godlovesme @kc_kennylau

  16. kc_kennylau
    • one year ago
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    Well, you have the length of \(CD\) and \(CB\)

  17. Liv1234
    • one year ago
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    So, 30*17?

  18. kc_kennylau
    • one year ago
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    Well, the formula is \(CD^2 = CB\times CA\)

  19. Liv1234
    • one year ago
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    Ohh, so 30^2=17*CA?

  20. kc_kennylau
    • one year ago
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    Yes :DDDD

  21. Liv1234
    • one year ago
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    A!

  22. Liv1234
    • one year ago
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    Can you help me with another question(:

  23. kc_kennylau
    • one year ago
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    No it's not A

  24. kc_kennylau
    • one year ago
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    A is actually the length of \(CA\)

  25. kc_kennylau
    • one year ago
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    You're asked to find the length of \(BA\)

  26. Liv1234
    • one year ago
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    Yes, it is because 30^2=900 and 17*52.9=900, right?

  27. Liv1234
    • one year ago
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    Ohh I see what you're saying sorry my bad.

  28. kc_kennylau
    • one year ago
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    it's alright xd

  29. Liv1234
    • one year ago
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    So, where did I mess up though?

  30. kc_kennylau
    • one year ago
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    Well, you only found the length of \(CA\).

  31. kc_kennylau
    • one year ago
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    How do you find the length of \(BA\) from that?

  32. Liv1234
    • one year ago
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    Divide by half?

  33. kc_kennylau
    • one year ago
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    Well, we can see that \(CA - CB = BA\)

  34. Liv1234
    • one year ago
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    Ohh, so 30-17?

  35. kc_kennylau
    • one year ago
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    Um.... no

  36. kc_kennylau
    • one year ago
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    Remember that \(CA = 52.9\)

  37. Liv1234
    • one year ago
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    Ohhh okay oopsies. cx

  38. Liv1234
    • one year ago
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    It's 35.9! Because 52.9-17=35.9

  39. kc_kennylau
    • one year ago
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    :D

  40. kc_kennylau
    • one year ago
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    yes :D

  41. Liv1234
    • one year ago
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    Can you help me again??

  42. kc_kennylau
    • one year ago
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    of course

  43. Liv1234
    • one year ago
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  44. kc_kennylau
    • one year ago
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    Well, can you tell me \(m\angle OAB\)?

  45. Liv1234
    • one year ago
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    No, I don't understand it. :/ (and by the way I need help with like 3 other questions after this one if that's okay)

  46. kc_kennylau
    • one year ago
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    I mean the angle \(OAB\)

  47. Liv1234
    • one year ago
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    I don't know how to find it out that's what I'm confused about.

  48. kc_kennylau
    • one year ago
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    Well, since \(AB\) is tangent, the angle is actually \(90^\circ\)

  49. Liv1234
    • one year ago
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    Okay, so what do we do from there?

  50. kc_kennylau
    • one year ago
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    So, we can actually apply the Pythagoras' Theorem

  51. Liv1234
    • one year ago
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    Ohh okay. So, what was the formula for that?

  52. kc_kennylau
    • one year ago
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    \(OA^2+AB^2=OB^2\)

  53. Liv1234
    • one year ago
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    So, 24^2+AB^2=OB^2?

  54. Liv1234
    • one year ago
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    Did I put the numbers in wrong?

  55. kc_kennylau
    • one year ago
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    Nope

  56. kc_kennylau
    • one year ago
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    Can we find \(OC\)?

  57. Liv1234
    • one year ago
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    I think we can, but I'm not sure how.

  58. kc_kennylau
    • one year ago
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    Well, it is another radius

  59. kc_kennylau
    • one year ago
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    All radius have the same length

  60. Liv1234
    • one year ago
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    So, what would it be then?

  61. Liv1234
    • one year ago
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    I mean the radius.

  62. Liv1234
    • one year ago
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    Hello?

  63. kc_kennylau
    • one year ago
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    Well

  64. kc_kennylau
    • one year ago
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    \(OA\) is another radius

  65. Liv1234
    • one year ago
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    Ohh, so 90?

  66. kc_kennylau
    • one year ago
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    Em... what are we talking about?

  67. Liv1234
    • one year ago
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    The radius?

  68. kc_kennylau
    • one year ago
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    Well, the radius is 24

  69. Liv1234
    • one year ago
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    Oh okay, so OC would be 24?

  70. kc_kennylau
    • one year ago
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    Yes :D

  71. kc_kennylau
    • one year ago
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    So how long is \(OB\)?

  72. Liv1234
    • one year ago
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    24!

  73. kc_kennylau
    • one year ago
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    em.. i'm talking about \(OB\)

  74. Liv1234
    • one year ago
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    Ohh, oops.

  75. Liv1234
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    I'm not sure what it would be, is there a way to find out?

  76. kc_kennylau
    • one year ago
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    Well, OB is just OC + CB

  77. Liv1234
    • one year ago
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    Would it be 48?

  78. kc_kennylau
    • one year ago
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    Nope. CB is actually 27

  79. Liv1234
    • one year ago
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    27+24=51

  80. Liv1234
    • one year ago
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    So, would the answer to the question be 51?

  81. kc_kennylau
    • one year ago
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    Nope, we need to find \(AB\)

  82. Liv1234
    • one year ago
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    And we'd do that by?

  83. kc_kennylau
    • one year ago
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    the pythagoras' theorem! :D

  84. Liv1234
    • one year ago
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    Ohh. And what was the formula again aha?

  85. kc_kennylau
    • one year ago
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    \(OA^2+AB^2=OB^2\)

  86. Liv1234
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    So, 27^2+AB^2=51^2. 51^2=2601. Right?

  87. Liv1234
    • one year ago
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    I might have messed up with that a little.

  88. kc_kennylau
    • one year ago
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    it should be 24^2 instead of 27^2

  89. Liv1234
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    Ohh okay. 24^2+AB^2=51^2.

  90. kc_kennylau
    • one year ago
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    :)

  91. Liv1234
    • one year ago
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    It would be 45. Which is A. Can you help me again?(:

  92. Liv1234
    • one year ago
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  93. kc_kennylau
    • one year ago
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    Well, it's difficult to discuss without names

  94. Liv1234
    • one year ago
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    What do you mean?

  95. kc_kennylau
    • one year ago
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    Without names of vertices

  96. kc_kennylau
    • one year ago
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    I can't refer any line

  97. Liv1234
    • one year ago
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    Want me to post another question instead?

  98. kc_kennylau
    • one year ago
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    no

  99. kc_kennylau
    • one year ago
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    |dw:1434224619983:dw|

  100. Liv1234
    • one year ago
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    Oh okay, that's gonna make it a lot easier.

  101. kc_kennylau
    • one year ago
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    So, O is the centre of the circle

  102. kc_kennylau
    • one year ago
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    And OM makes a right angle with AB

  103. kc_kennylau
    • one year ago
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    So, AM=MB

  104. kc_kennylau
    • one year ago
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    (Just remember this result :p)

  105. Liv1234
    • one year ago
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    Okay. I will.

  106. Liv1234
    • one year ago
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    So, what do we start out with?

  107. Liv1234
    • one year ago
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    I'm confused.

  108. kc_kennylau
    • one year ago
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    So AM=MB

  109. Liv1234
    • one year ago
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    So, 21=5?

  110. Liv1234
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    I mean 5=21?

  111. kc_kennylau
    • one year ago
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    Em...

  112. kc_kennylau
    • one year ago
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    I mean that M divides AB in half

  113. Liv1234
    • one year ago
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    So, half of 21?

  114. kc_kennylau
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    yes, that would be MB

  115. Liv1234
    • one year ago
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    And half of 21 is 10.5

  116. kc_kennylau
    • one year ago
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    |dw:1434225065530:dw|

  117. kc_kennylau
    • one year ago
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    Yes

  118. Liv1234
    • one year ago
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    So, AM=10.5?

  119. Liv1234
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    Right?

  120. kc_kennylau
    • one year ago
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    yes

  121. Liv1234
    • one year ago
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    Then what would we do?

  122. kc_kennylau
    • one year ago
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    Apply the Pythagoras' Theorem

  123. Liv1234
    • one year ago
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    Can you write it out for me again please?

  124. kc_kennylau
    • one year ago
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    \(OM^2+MB^2=OB^2\)

  125. Liv1234
    • one year ago
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    OM^2+10.5^2=OB^2?

  126. kc_kennylau
    • one year ago
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    You know OM

  127. Liv1234
    • one year ago
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    I do?

  128. kc_kennylau
    • one year ago
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    It's 5

  129. Liv1234
    • one year ago
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    5^2+10.5^2=OB^2?

  130. kc_kennylau
    • one year ago
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    yes

  131. Liv1234
    • one year ago
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    I got 135.25^2

  132. kc_kennylau
    • one year ago
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    yes

  133. Liv1234
    • one year ago
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    Which is 18,300.

  134. kc_kennylau
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    that is OB^2

  135. kc_kennylau
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    No, OB^2 is 135.25

  136. Liv1234
    • one year ago
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    I know, but when you do 135.25^2 you get 18300(rounded)

  137. kc_kennylau
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    No, OB^2 is 135.25, so you root it, not square it

  138. Liv1234
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    Okay.

  139. Liv1234
    • one year ago
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    Would it be A? 11.6?

  140. kc_kennylau
    • one year ago
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    yes

  141. Liv1234
    • one year ago
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  142. kc_kennylau
    • one year ago
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    This is a theorem

  143. Liv1234
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    What does that mean?

  144. Liv1234
    • one year ago
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    Can you help me with it?

  145. kc_kennylau
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    Just times two

  146. Liv1234
    • one year ago
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    48?

  147. kc_kennylau
    • one year ago
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    The angle at center \(m\angle O\) is twice the angle at circumference \(m\angle R\) yes

  148. Liv1234
    • one year ago
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  149. Liv1234
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    What about this one?

  150. Liv1234
    • one year ago
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    And this is the last one.

  151. kc_kennylau
    • one year ago
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    Secant secant theorem: \(CA\times CB = CE \times CD\)

  152. Liv1234
    • one year ago
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    42*18=4*CD?

  153. kc_kennylau
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    CA isn't 42

  154. kc_kennylau
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    4 is CD, not CE

  155. Liv1234
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    Okay, so 18*CB=CE*4?

  156. Liv1234
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    CA*18=CE*4?

  157. kc_kennylau
    • one year ago
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    Yeah, how to find CA? It's the sum of two lines again

  158. Liv1234
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    CA=64?

  159. Liv1234
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    64*18=CE*4?

  160. kc_kennylau
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    not 64, it's 60

  161. Liv1234
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    60*18=CE*4?

  162. kc_kennylau
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    yes

  163. Liv1234
    • one year ago
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    So, 270? Which is D?

  164. Liv1234
    • one year ago
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    Right?

  165. kc_kennylau
    • one year ago
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    no

  166. kc_kennylau
    • one year ago
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    You're asked to find DE

  167. kc_kennylau
    • one year ago
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    You need to deduct 4

  168. Liv1234
    • one year ago
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    1,066?

  169. kc_kennylau
    • one year ago
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    266

  170. kc_kennylau
    • one year ago
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    270-4=266

  171. Liv1234
    • one year ago
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    Ohh, that makes sense, so C?

  172. Liv1234
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    ?

  173. kc_kennylau
    • one year ago
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    yes

  174. Liv1234
    • one year ago
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    YAY thank you so much for helping(:

  175. kc_kennylau
    • one year ago
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    no problem

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