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.
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AonZ
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AonZ
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@jim_thompson5910 @Mertsj help pls
AonZ
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umm im pretty sure ADC is 180. Why wouldnt it be?
Loser66
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since I saw (theta) symbol at ADB and you indicate about ADC is theta
AonZ
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im sorry ADB is theta :P
Loser66
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ok, let tag someone who is good at geometry and trig @Mertsj
Mertsj
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AonZ
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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
Mertsj
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\[c^2=m^2+(\frac{b}{2})^2-2m(\frac{b}{2})\cos \theta \]
Mertsj
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Solve for cos theta
AonZ
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ok yea i proved that part
AonZ
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and cos(180-theta)
is putting a negative sign infront of that equation
AonZ
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how can we prove part b though?
Mertsj
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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
AonZ
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- 4m^2 + b^2 - 4c^2 / 4mb = 4m^2 + b^2 - 4a^2 / 4mb
right?
AonZ
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ok i solved it thank you