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If someone can help, i'd really appreciate it. Thanks

- Thread starter aek
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- #1

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If someone can help, i'd really appreciate it. Thanks

- #2

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How have you attempted to prove this statement ?aek said:

If someone can help, i'd really appreciate it. Thanks

- #3

Curious3141

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HINT : Apply cosine rule on triangle PQA.aek said:

If someone can help, i'd really appreciate it. Thanks

- #4

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Why are you answering my question with a question? Just abit ironic thats all. Thanks for the hint hyper, it was helpful.

- #5

Curious3141

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I can't even tell if that's sarcasm. In any case, it was I that gave the hint, not hyper.aek said:Why are you answering my question with a question? Just abit ironic thats all. Thanks for the hint hyper, it was helpful.

The policy of this forum is not to do homework or just solve the problem for the poster unless it is evident some effort or at least some serious thinking has been put into the solution by the poster. So I can understand why hyper posted what he did.

Have you solved the problem yet ? If you're still finding difficulty, apply the Cosine Rule as I suggested and see what you get. If stuck, post here with working, then you'll get help.

- #6

Icebreaker

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yep i got the point,

but i too know that you have to use the cosine rule and i've tried and couldn't find a relevant answer. If you want proof that i tried, i could scan my work but you being so trustful that wouldn't be necassary :P

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- #9

Curious3141

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[tex]PQ = QR = 3QA[/tex]aek said:yep i got the point,

but i too know that you have to use the cosine rule and i've tried and couldn't find a relevant answer. If you want proof that i tried, i could scan my work but you being so trustful that wouldn't be necassary :P

[tex]PA^2 = PQ^2 + QA^2 - 2(PQ)(QA)\cos 60^o[/tex]

[tex]PA^2 = 9QA^2 + QA^2 - 2(3QA)(QA)(\frac{1}{2})[/tex]

[tex]PA^2 = 7QA^2[/tex]

Perhaps now the sarcasm can end.

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