anonymous
  • anonymous
In the diagram, triangle ABC is right-angled and D is the mid-point of BC. Angle DAC=30 degrees and angle BAD =x degrees, denoting the length of AD by l, (i) express each of AC and BC in terms of l, and show that AB= (1/2)l sqrt 7 (ii) show that x=tan^-1 (2 over sqrt 3) -30.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
|dw:1332840732002:dw|
anonymous
  • anonymous
@dpaInc
anonymous
  • anonymous
i) using the properties of a 30,60,90 triangle. AC = 1/2*I*sqrt(3) BC = 2*(I/2) = I

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anonymous
  • anonymous
and AB?
anonymous
  • anonymous
AB^2 = AC^2 + BC^2 Pythagorean theorem
anonymous
  • anonymous
So, how would you plug in the values for that? And how about part (ii)?
anonymous
  • anonymous
You have AB and BC, so add the squares, then get the root of the sum.
anonymous
  • anonymous
BC = 2*(I/2) or BC = 2*(I/2)=l???
anonymous
  • anonymous
|dw:1332841919210:dw|
anonymous
  • anonymous
OK.
anonymous
  • anonymous
|dw:1332842169543:dw| you can solve that...
anonymous
  • anonymous
I still don't know AB...?
anonymous
  • anonymous
\[2*I/2 = I\] The 2s cancel each other So you have: \[AB^2 = AC^2+BC^2\] Where, \[AC =I\sqrt{3}/2\] \[BC = I\] \[AB^2 = (I\sqrt{3}/2)^2 + I^2\] Can you simplify this and find the root to find AB?
anonymous
  • anonymous
YEs
anonymous
  • anonymous
actually, @mcnalljj , how do you simplify that?
anonymous
  • anonymous
Maybe it is better set as: \[AB^2 = (\sqrt{3}/2*I) +I^2\] \[AB^2 = I^2 + (3/4*I)\]
anonymous
  • anonymous
How do you carry on from here?
anonymous
  • anonymous
I'm not certain if it can be much further simplified: \[AB = \sqrt{I(I+3/4)}\] Might need checking
anonymous
  • anonymous
Well, the answer should be = to (1/2)l sqrt 7
anonymous
  • anonymous
@mcnalljj You're correct all along, just miscalculation at: AB² = (3/4) I² + I² = ( 7/4) I² =>AB = (√ 7/2) I
anonymous
  • anonymous
@order You need to learn calculate with the solvers!
anonymous
  • anonymous
Ah OK. thanks :)
anonymous
  • anonymous
Thanks, I thought I'd messed it all up :) was going to re-do it when I had a chance. Can't say that I have been practicing anything like this in the last...5+years :S
Abmon98
  • Abmon98
Can someone please help me I that question I'm stuck on it

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