harder 3unit question (1 Viewer)

[conics2008]

help needed please thanks =]

 

[Trebla]

Let < C₂AC₁ = x
Since ΔC₂AC₁ is isosceles (from two equal sides), then
< AC₂C₁ = π/2 - x/2
Hence < AC₂B = π - (π/2 - x/2)
In ΔAC₂B:
sin B / b = sin {π - (π/2 - x/2)} / c
=> sin B / b = sin (π/2 - x/2) / c
=> sin B / b = cos (x/2) / c
=> cos (x/2) = c.sin B / b
=> cos²(x/2) = c²sin²B / b²
Also,
(a₁ - a₂)² = 2b² - 2b²cos x
= 2b² - 2b²(2cos²(x/2) - 1)
= 2b² - 2b²(2c²sin²B / b² - 1)
= 4b² - 4c²sin²B
= 4(b² - c²sin²B)
Hence a₁ - a₂ = 2√(b² - c²sin²B) taking the positive root as a₁ - a₂ > 0.

 

[Michaelmoo]

Let < C₂AC₁ = x
Since ΔC₂AC₁ is isosceles (from two equal sides), then
< AC₂C₁ = π/2 - x/2
Hence < AC₂B = π - (π/2 - x/2)
In ΔAC₂B:
sin B / b = sin {π - (π/2 - x/2)} / c
=> sin B / b = sin (π/2 - x/2) / c
=> sin B / b = cos (x/2) / c
=> cos (x/2) = c.sin B / b
=> cos²(x/2) = c²sin²B / b²
Also,
(a₁ - a₂)² = 2b² - 2b²cos x
= 2b² - 2b²(2cos²(x/2) - 1)
= 2b² - 2b²(2c²sin²B / b² - 1)
= 4b² - 4c²sin²B
= 4(b² - c²sin²B)
Hence a₁ - a₂ = 2√(b² - c²sin²B) taking the positive root as a₁ - a₂ > 0.

Hey. Can I ask, I understand why you did all that work after "Also", But what made you do the work before that??? Why did you go that path?

Thanks.

 

[Trebla]

Hey. Can I ask, I understand why you did all that work after "Also", But what made you do the work before that??? Why did you go that path?

Thanks.

I chose that path because it works!!! lol
Nah, think about it. How else can one obtain the 'sin B' in the formula?

 

[conics2008]

i was going with cosine rule and everything...

Thanks terbla

 

[Trebla]

O and Btw

sin B / b = sin {π - (π/2 - x/2)} / c
=> sin B / b = sin (π/2 - x/2) / c


isnt the second line supposed to be

sinB / b = sin (n/2 + x/2) / c

I actually applied the theorem: sin (π - x) = sin x, rather than simplifying it completely, because it's easier to gain the cosine function that way...
And yes: sin (n/2 + x/2) = sin (π/2 - x/2)
You can verify this by expanding it.