trigonometry maths a level question

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Qxi.xli
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im struggling to see where what I did was wrong? I mean, i did get a negative answer lol but my reasoning seems fine? if that makes sense.
the mark scheme did it differently to me. any help would be appreciated. thanks Name:  WhatsApp Image 2021-01-20 at 11.35.38 AM.jpeg
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mqb2766
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Can you explain your approach a bit more? It should be reasonably straightforward?
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DFranklin
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I can't see how your working relates to the question - are you sure you've posted the right question?
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Qxi.xli
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(Original post by mqb2766)
Can you explain your approach a bit more? It should be reasonably straightforward?
i used the sine rule to get two equations in terms of h and x. i then rearranged to get both of them in terms of x, then equated the then re-arranged to find the value of h?
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mqb2766
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(Original post by vix.xvi)
i used the sine rule to get two equations in terms of h and x. i then rearranged to get both of them in terms of x, then equated the then re-arranged to find the value of h?
But you don't seem to use 80m, why 32 and 38 degrees ....
Rather than the trig mechanics, what are you trying to find and how?
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DFranklin
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Note that the OP writes 12a for the question they're answering, but the attachment is for a question 6...
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Qxi.xli
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(Original post by DFranklin)
Note that the OP writes 12a for the question they're answering, but the attachment is for a question 6...
Omg tysm for pointing that out

(Original post by mqb2766)
But you don't seem to use 80m, why 32 and 38 degrees ....
Rather than the trig mechanics, what are you trying to find and how?
I attached the wrong photo im so sorry. Here's the correct question. Name:  IMG_20210120_155350903.jpg
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Muttley79
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(Original post by vix.xvi)
Omg tysm for pointing that out


I attached the wrong photo im so sorry. Here's the correct question. Name:  IMG_20210120_155350903.jpg
Views: 9
Size:  122.4 KB
You've gone wrong when you multiply through as you've 'ignored' doing anything to the 15.

You seem to have made it very complicated - they are right angled triangles so you could have used tan

e.g. tan 40 = h/(x + 15) and ...
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DFranklin
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(Original post by vix.xvi)
Omg tysm for pointing that out
Your 50^o angle seems to change to 30^o for no obvious reason.

Your workings would be a lot simpler if you used cos <angle> instead of manually converting to sin(90 - <angle>) and you'd be less likely to make mistakes.

Even better, you could directly use tan instead of sin(angle)/sin(90-angle) as you are doing.
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DJ Trump fan
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Just split it into a right angle triangle and a non right angle triangle. Work out the hypotenuese of the right angled triangle by using sine rule for the non right angled triangle. You can then just use SOHCAHTOA to find the height. As someone said above, you haven't multiplied the 15 by sin52 in your working out when you rearrange. Hope this helps.
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Qxi.xli
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(Original post by DFranklin)
Your 50^o angle seems to change to 30^o for no obvious reason.

Your workings would be a lot simpler if you used cos <angle> instead of manually converting to sin(90 - <angle>) and you'd be less likely to make mistakes.

Even better, you could directly use tan instead of sin(angle)/sin(90-angle) as you are doing.
(Original post by Muttley79)
You've gone wrong when you multiply through as you've 'ignored' doing anything to the 15.

You seem to have made it very complicated - they are right angled triangles so you could have used tan

e.g. tan 40 = h/(x + 15) and ...
Thank you for your help x . I tried it using tan it was way easier. Thanks
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Muttley79
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(Original post by vix.xvi)
Thank you for your help x . I tried it using tan it was way easier. Thanks
Yes - only use sine/cosine rule when there is no right angle
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