# Maths trigonometry

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#1
PQR is a right angle triangle. QS is perpendicular to PR, and PQ has length 1.

1) Find QS and PS in-terms of 0.

2) By considering triangle QSR, find an expression for SR.

3) By considering triangle PQR, find PR in-terms of e.

4) Hence, show that
cos A = sec 0 -sin o tan 0.

Last edited by Omarblack; 1 month ago
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1 month ago
#2
(Original post by Omarblack)
PQR is a right angle triangle. QS is perpendicular to PR, and PQ has length 1.

1) Find QS and PS in-terms of 0.

2) By considering triangle QSR, find an expression for SR.

3) By considering triangle PQR, find PR in-terms of e.

4) Hence, show that
cos A = sec 0 -sin o tan 0.
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#3
Here is my workings out so far but not sure if it's right and I'm stuck.
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#4
(Original post by Muttley79)
I have.
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1 month ago
#5
(Original post by Omarblack)
I have.
OK I'll take a look - this wasn't there when I replied
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#6
(Original post by Muttley79)
OK I'll take a look - this wasn't there when I replied
OK, thank you very much
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1 month ago
#7
(Original post by Omarblack)
OK, thank you very much
OK I go the result - you need to rewrite what you have:

QS =

PS =

SR =

PR = PS + SR .... then rearrange
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#8
What do you mean by rewrite?

Do you mean make it equal to PS and RS and SR in the separate questions?

What do you mean by rewrite?

Do you mean make it equal to PS and RS and SR in the separate questions?
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1 month ago
#9
(Original post by Omarblack)
What do you mean by rewrite?

Do you mean make it equal to PS and RS and SR in the separate questions?

What do you mean by rewrite?

Do you mean make it equal to PS and RS and SR in the separate questions?
You need to quote me so I can see you've responded.

Follow the question through as it asks. You get expressions for all these lengths.

Then put them together at the end: expression for PR = PS + SR

Write out as I suggested:

sin theta = QS/1 so QS =

tan theta =

No need to use Pythag as we know angle SQR = theta as well
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#10
(Original post by Muttley79)
OK I go the result - you need to rewrite what you have:

QS =

PS =

SR =

PR = PS + SR .... then rearrange

This is what I've got?
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1 month ago
#11
(Original post by Omarblack)
This is what I've got?
See my hint about QS ...
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#12
(Original post by Muttley79)
You need to quote me so I can see you've responded.

Follow the question through as it asks. You get expressions for all these lengths.

Then put them together at the end: expression for PR = PS + SR

Write out as I suggested:

sin theta = QS/1 so QS =

tan theta =

No need to use Pythag as we know angle SQR = theta as well
(Original post by Muttley79)
See my hint about QS ...
I think I'm following, so what you mean is since

sine (pheta) = 1/QS Then QS = tan(pheta) but after that point I'm lost because surely PS = QS / tan (pheta) so doesn't that mean QS would rearrange to PStan(pheta)
Also tan is sine/cos. So I'm really confused now. If we was following the the expressions and rearranging as we go then I must be doing something wrong in my workings out, but I don't know what it is.
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1 month ago
#13
(Original post by Omarblack)
I think I'm following, so what you mean is since

sine (pheta) = 1/QS Then QS = tan(pheta) but after that point I'm lost because surely PS = QS / tan (pheta) so doesn't that mean QS would rearrange to PStan(pheta)
Also tan is sine/cos. So I'm really confused now. If we was following the the expressions and rearranging as we go then I must be doing something wrong in my workings out, but I don't know what it is.
sin theta = opp/hyp = QS/1

tan theta = QS/PS so PS =QS/tan theta now sub in QS

Sorry back later ... need to go
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#14
(Original post by Muttley79)
sin theta = opp/hyp = QS/1

tan theta = QS/PS so PS =QS/tan theta now sub in QS

Sorry back later ... need to go
When you see this, I think I understand.

Since sine(pheta) = QS/1 THEN QS = SINE(PHETA)

and PS = QS/TAN(PHETA).

I'M STILL PUZZLED ON THE REST THOUGH.
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1 month ago
#15
(Original post by Omarblack)
When you see this, I think I understand.

Since sine(pheta) = QS/1 THEN QS = SINE(PHETA)

and PS = QS/TAN(PHETA).

I'M STILL PUZZLED ON THE REST THOUGH.
Sub in QS and the start on the second triangle ... post what you get .. don't give up
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#16
(Original post by Muttley79)
Sub in QS and the start on the second triangle ... post what you get .. don't give up
OK thanks for the guidance so far. This is my current results for the expression for SR.

Is this correct? If it is could you give me some guidance for the whole triangle please?

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1 month ago
#17
(Original post by Omarblack)
OK thanks for the guidance so far. This is my current results for the expression for SR.

Is this correct? If it is could you give me some guidance for the whole triangle please?
Now put it all together:
PR = PS + SR

What did you have for PR and PS? I agree with SR
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#18
(Original post by Muttley79)
Now put it all together:
PR = PS + SR

What did you have for PR and PS? I agree with SR
I get QS/tan(pheta) + sin(pheta) tan(pheta) is this correct?
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1 month ago
#19
(Original post by Omarblack)
I get QS/tan(pheta) + sin(pheta) tan(pheta) is this correct?
You know QS = sin theta so simplify that.

Now put = PR
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#20
(Original post by Muttley79)
You know QS = sin theta so simplify that.

Now put = PR
After Simplification I have cos(pheta) sin(pheta) + sim(pheta) tan(pheta)?
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