AonZ
  • AonZ
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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AonZ
  • AonZ
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AonZ
  • AonZ
@jim_thompson5910 @Mertsj help pls
AonZ
  • AonZ
umm im pretty sure ADC is 180. Why wouldnt it be?

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Loser66
  • Loser66
since I saw (theta) symbol at ADB and you indicate about ADC is theta
AonZ
  • AonZ
im sorry ADB is theta :P
Loser66
  • Loser66
ok, let tag someone who is good at geometry and trig @Mertsj
Mertsj
  • Mertsj
|dw:1367623308982:dw|
AonZ
  • AonZ
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
Mertsj
  • Mertsj
\[c^2=m^2+(\frac{b}{2})^2-2m(\frac{b}{2})\cos \theta \]
Mertsj
  • Mertsj
Solve for cos theta
AonZ
  • AonZ
ok yea i proved that part
AonZ
  • AonZ
and cos(180-theta) is putting a negative sign infront of that equation
AonZ
  • AonZ
how can we prove part b though?
Mertsj
  • Mertsj
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
AonZ
  • AonZ
- 4m^2 + b^2 - 4c^2 / 4mb = 4m^2 + b^2 - 4a^2 / 4mb right?
AonZ
  • AonZ
ok i solved it thank you

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