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AonZ

  • 3 years ago

Yr11 3unit question help pls! ABC is a triangle and D is the midpoint of AC. BD=m and angle ADB = theta (a) show that cos(theta) = 4m^2 + b^2 - 4c^2 / 4mb and write an expression for cos(180-theta). (b) show a^2 +c^2 = 2m^2 + 1/2 b^2 I have done part (a) and need help on part (b) pls.

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  1. AonZ
    • 3 years ago
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    |dw:1367620733749:dw|

  2. AonZ
    • 3 years ago
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    @jim_thompson5910 @Mertsj help pls

  3. AonZ
    • 3 years ago
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    umm im pretty sure ADC is 180. Why wouldnt it be?

  4. Loser66
    • 3 years ago
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    since I saw (theta) symbol at ADB and you indicate about ADC is theta

  5. AonZ
    • 3 years ago
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    im sorry ADB is theta :P

  6. Loser66
    • 3 years ago
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    ok, let tag someone who is good at geometry and trig @Mertsj

  7. Mertsj
    • 3 years ago
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    |dw:1367623308982:dw|

  8. AonZ
    • 3 years ago
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    hmm yes i proved part a but i couldnt get part b. I think you need to do something with part a to get part b

  9. Mertsj
    • 3 years ago
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    \[c^2=m^2+(\frac{b}{2})^2-2m(\frac{b}{2})\cos \theta \]

  10. Mertsj
    • 3 years ago
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    Solve for cos theta

  11. AonZ
    • 3 years ago
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    ok yea i proved that part

  12. AonZ
    • 3 years ago
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    and cos(180-theta) is putting a negative sign infront of that equation

  13. AonZ
    • 3 years ago
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    how can we prove part b though?

  14. Mertsj
    • 3 years ago
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    From the first part you have an expression for cos (180-theta) Get another expression for cos (180-theta) using triangle BCD Equate those two expressions for cos(180-theta) and solve for a^2+b^2

  15. AonZ
    • 3 years ago
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    - 4m^2 + b^2 - 4c^2 / 4mb = 4m^2 + b^2 - 4a^2 / 4mb right?

  16. AonZ
    • 3 years ago
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    ok i solved it thank you

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