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# Missing angle help watch

1. This is kind of related to physics but it is mainly maths so i'll post it here. I have worked out the critical angle to be 75.2 degrees (if that has any relevance?!)

How do you work out angle i? It is only worth one mark so i assume its quite simple but nevertheless, I'm a bit confused. Thanks

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2. (Original post by jessyjellytot14)
This is kind of related to physics but it is mainly maths so i'll post it here. I have worked out the critical angle to be 75.2 degrees (if that has any relevance?!)

How do you work out angle i? It is only worth one mark so i assume its quite simple but nevertheless, I'm a bit confused. Thanks

Posted from TSR Mobile
You should know that, by the law of reflection, the angle of incidence = the angle of reflection. Apply that to the 85 degrees and you will get the top left angle in the triangle. Then it is just subtraction from 180 degrees to get the required angle
3. (Original post by PhyM23)
You should know that, by the law of reflection, the angle of incidence = the angle of reflection. Apply that to the 85 degrees and you will get the top left angle in the triangle. Then it is just subtraction from 180 degrees to get the required angle
So would angle i and angle r both be 65 degrees?
4. (Original post by jessyjellytot14)
So would angle i and angle r both be 65 degrees?
Almost - you are correct in saying that angle i = 65 degrees, but at this stage you will need to use Snell's Law [n1sin(theta1) = n2sin(theta2)] in order to get r, since refraction, not reflection, is taking place at this point

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