Qer
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#1
Report Thread starter 4 years ago
#1
http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf

I totally don't understand part c of question 6
finding AC

i think that we can use sine rule here like that
9/sin2.04 =ac/sin 0.7 { ( 0.7 + pi/2 )-pi=0.7 radian} ( is that right way)
and i got 3.09.....

but in mark scheme, they use tan 0.7=ac/9


is my method is valid
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Swissblade
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#2
Report 4 years ago
#2
(Original post by Qer)
http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf

I totally don't understand part c of question 6
finding AC

i think that we can use sine rule here like that
9/sin2.04 =ac/sin 0.7 { ( 0.7 + pi/2 )-pi=0.7 radian} ( is that right way)
and i got 3.09.....

but in mark scheme, they use tan 0.7=ac/9


is my method is valid
OA is perpendicular to AC - there's a 90° angle. Hence you only need to use sin/cos/tan, not sine or cosine rule (you use this when there isn't a 90° angle).
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Qer
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#3
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#3
(Original post by markschemes)
oa is perpendicular to ac - there's a 90 degree angle. Hence you only need to use sin/cos/tan, not sine or cosine rule.
can you please explain me ? I didn't understand
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Qer
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#4
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#4
(Original post by markschemes)
OA is perpendicular to AC - there's a 90° angle. Hence you only need to use sin/cos/tan, not sine or cosine rule (you use this when there isn't a 90° angle).
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Philip-flop
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#5
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#5
(Original post by Qer)
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What seems to be your problem?

Do you not understand Pythagoras?
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RDKGames
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#6
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#6
(Original post by Qer)
http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf

I totally don't understand part c of question 6
finding AC

i think that we can use sine rule here like that
9/sin2.04 =ac/sin 0.7 { ( 0.7 + pi/2 )-pi=0.7 radian} ( is that right way)
and i got 3.09.....

but in mark scheme, they use tan 0.7=ac/9


is my method is valid
It's a right angled triangle... you don't need to use the sine rule...... you only need to use basic trigonometry...

BUT if you insist on doing so, then to find length you need to do \displaystyle \frac{9}{\sin(\frac{\pi}{2}-0.7)}=\frac{AC}{\sin(0.7)} which gives AC=7.58
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Swissblade
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#7
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#7
(Original post by Qer)
???
Where's the confusion? The text says only use sin/cos/tan if there's a 90 degree angle.

For the question in OP, tan(x) = \frac{opposite}{adjacent} \Rightarrow tan(0.7) = \frac{opposite}{9}, rearrange for opposite. This is a GCSE topic buddy.
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RDKGames
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#8
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#8
(Original post by Qer)
???
What you have highlighted does not refer to the sine rule which is \frac{\sin(A)}{a}=\frac{\sin(B)}{b}=\frac{\sin(C)}{c}

nor the cosine rule which is a^2=b^2+c^2-2bc\cos(A)
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Swissblade
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#9
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#9
(Original post by Qer)
i am just try to clear my confusion.the markscheme said that you can't use sine and cosine in right angle triangle when 90 degree angle is present.............while book says its better
You're confusing sine/cosine/tangent, which are functions, with the "sine rule" and "cosine rule", which are equations. They're not the same at all.
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Qer
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#10
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#10
(Original post by RDKGames)
It's a right angled triangle... you don't need to use the sine rule...... you only need to use basic trigonometry...

BUT if you insist on doing so, then to find length you need to do \displaystyle \frac{9}{\sin(\frac{\pi}{2}-0.7)}=\frac{AC}{\sin(0.7)} which gives AC=7.58
OK I understand


one thing still confuses me that why you use pi/2 -0.7

why we don't use pi -( pi/2 +0.7 )
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RDKGames
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#11
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#11
(Original post by Qer)
OK I understand


one thing still confuses me that why you use pi/2 -0.7

why we don't use pi -( pi/2 +0.7 )
It's the same thing
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Qer
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#12
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#12
(Original post by RDKGames)
It's the same thing
ohh

just confusing myself :afraid:

thanks
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Qer
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#13
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#13
thanks, everyone.
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