trigo
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from riverside prelim's 2008 paper 2:
hi guys, can someone please help me out with question b? i've got question a)'s answer(it's 25m) but my answer for is b)142.67m, which is not the same as the answer key. answer to question b is 95.4m.
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is it me or what,i see no peektures
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http://i38.tinypic.com/24zdezs.jpg here's the link!
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I got a different answer from you and the answer key.. HAHA, let me try again. Meanwhile anyone who can solve it, heh heh. ^^"
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sorry cant help oso..my ans is 133.6 which is wrong lol. wait for eagle or tintin to help!
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tried again then get 86.6. Wrong again! lmao...u sure ur ans sheet correct?
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Er for part b i got...,
AH^2=200^2+100^2-2(100)(200)cos30
and i got 123.931
Feel free to correct me if i am wrong.
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Originally posted by jlzk:
Er for part b i got...,
AH^2=200^2+100^2-2(100)(200)cos30
and i got 123.931
Feel free to correct me if i am wrong.
lol i oso got..i think the hard part to prove tht AT=95.4 tht one \:
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200^2 = 100^2 + x^2 - 2(100)(x)cos53.8degrees
solve for X = AH.
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Originally posted by Maith:
lol i oso got..i think the hard part to prove tht AT=95.4 tht one \:
i really dun quite get it wad TS asking is it for AH or to prove but if prove he will have ask how to prove ma ..so..ya.. x.x
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I work backwards,
AT = 95.4 (given)
GT = GH + HT = 50 +25 = 75 since GH = EC = 25 cm
AT^2 = AG^2 + GT^2
95.4.^2 = AG^2 + 75^2
AG = 58.96 cm
AH^2 = AG^2 + GH^2 = 58.96^2 + 25^
AH = 64 cm leh
When I work forwards,
I get the same answer as jlzk ie 123.93 cm leh.
So, does this mean that there is something wrong with the question huh ?
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Originally posted by Lee012lee:
I work backwards,
AT = 95.4 (given)
GT = GH + HT = 50 +25 = 75 since GH = EC = 25 cm
AT^2 = AG^2 + GT^2
95.4.^2 = AG^2 + 75^2
AG = 58.96 cm
AH^2 = AG^2 + GH^2 = 58.96^2 + 25^
AH = 64 cm leh
When I work forwards,
I get the same answer as jlzk ie 123.93 cm leh.
So, does this mean that there is something wrong with the question huh ?
did i did it wrongly?cause i tot u take 90-60=30 thats why get 30 .Cause first part i also treat DCB as right angle to find ..so i pressume the bottom also the same
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By using sine rule the answer will be different, but by using cosine rule you get the answer, interesting..
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Originally posted by iamapebble:
By using sine rule the answer will be different, but by using cosine rule you get the answer, interesting..
Really? Lols..i though a answer can obtain as long the method is correct ..
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Originally posted by jlzk:
Really? Lols..i though a answer can obtain as long the method is correct ..
Yup, for o'levels yes. But for school papers sometimes the setter may give the answer a value that make the question logical, but end up neglecting one little part as the answer may vary or something..
But for this question, I'm not sure if it's this case or maybe cosine rule is the only method you can apply for this case? Let's wait for others to confirm :) -
what's e answer for AH?
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Originally posted by tinuviel07:
what's e answer for AH?
Not sure leh,
I work backwards AH = 64 cm but I work forwards AH = 123.93 leh, same as jlkz..
May be question something wrong lah.
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hmmm okie.. shall slowly read since i don have a calculator with me so got to use excel
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for question b, confirm AT = 143 m (3 s.f.)
Used 2 methods to confirm.
It is definitely not 95.4 m
But then, if you don't consider the 30 degrees,
AH² + 100² - 2(AH)(100) cos 53.8 = 200²
AH² -120AH – 30000 = 0
AH = (120 +- sqrt(120² +4*30000))/2
AH = 243.3m or -123m (N.A.) -
Originally posted by eagle:
for question b, confirm AT = 143 m (3 s.f.)
Used 2 methods to confirm.
It is definitely not 95.4 m
Trust yourself when the question has mistakes... :D
lol i thought i wrong.. yay
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I did it another method got a different AH... Hmmm
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my AH is
252.1749281 shrugs..
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I try doing some calculations to change the question properly...
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Try changing
HT is a vertical lower where its height is 50 m and ∡AHB = 53.8°.
to
HT is a vertical lower where its height is 50 m and ∡HAB = 51.4°.
AB change to 126.5 m
then can solve LOL
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thanks guys! i was racking my brain trying to figure out the answer -_-