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HSC 2012 MX1 Marathon #2 (archive) (1 Viewer)

barbernator

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Re: HSC 2012 Marathon :)

A man 1.8m tall, standing due east of a street light notices that his shadow is 6.5m long. He walks on a bearing of 150 deg T from his position for a distance of 12 m and now finds his shadow is 9.1m long. Find the height of the street light above the ground level. (in m correct to 1 dp)
 

Timske

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Re: HSC 2012 Marathon :)

A man 1.8m tall, standing due east of a street light notices that his shadow is 6.5m long. He walks on a bearing of 150 deg T from his position for a distance of 12 m and now finds his shadow is 9.1m long. Find the height of the street light above the ground level. (in m correct to 1 dp)
This is hard right?
 

Timske

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Re: HSC 2012 Marathon :)

just want to make sure im on the rite track

diagram.jpg
 

RishBonjour

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Re: HSC 2012 Marathon :)

I got 5.3 ?

my diagram is exactly like timske's

edit:
probably way off. haven't done trig in ages

whats zeee answerrrr.

Ill do the question again lol
 
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RishBonjour

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Re: HSC 2012 Marathon :)

I got the same diagram :/
ill try again later lol
 

Timske

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Re: HSC 2012 Marathon :)

Doing it now pretty hard
 

Timske

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Re: HSC 2012 Marathon :)

4.7 ?

Edit: 5.125
 
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Nooblet94

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Re: HSC 2012 Marathon :)

5.9?

Wouldn't be surprised if I'm wrong, my working's rushed and messy :p
 
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Timske

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Re: HSC 2012 Marathon :)

don't think im right either lol
 

Nooblet94

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Re: HSC 2012 Marathon :)

I'm pretty sure my method is correct... it's just a matter of whether or not I actually did the calculations correctly.
 

Timske

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Re: HSC 2012 Marathon :)

how didu do it
 

Timske

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Re: HSC 2012 Marathon :)

5.125 !!!!!
 

Nooblet94

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Re: HSC 2012 Marathon :)

I'll try describe it, but it's kind've tough without a diagram to point at and whatnot :p

(h=height of lamp, x=distance to lamp from pos. 1, y=distance to lamp from pos. 2)

I used the two pairs of similar triangles to create two expressions for h in terms of x and y and then equated the two. Rearranged that and found x in terms of y. Then I used the cosine rule in the 12/x/y triangle to get a quadratic in y. Solved for y and then subbed that value back into the expression for h.


(barb, i want teh answer, plz resp0nd)
 

RishBonjour

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Re: HSC 2012 Marathon :)

I used the cosine rule too. same method

I got 7.7 now..


Going to stop doing this, have bio assignment.
 

Timske

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Re: HSC 2012 Marathon :)

I think nooblet got it
 

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