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# C4 Trigonometry watch

1. http://img190.imageshack.us/img190/8396/img17946bt.jpg

sinα = 2/√5
cosα = 1/√5

sinß = 0.6
cosß = 0.8

Stuck on the last part! Any suggestions?
2. Without further geometry work you should be able to write as and hence use your known values to work out tan(LMN).

Edit: That is, i'm assuming you've done (iii) as you said 'last part'.
3. (Original post by dida87)
http://img190.imageshack.us/img190/8396/img17946bt.jpg

sinα = 2/√5
cosα = 1/√5

sinß = 0.6
cosß = 0.8

Stuck on the last part! Any suggestions?
get the intersections with the x-axis [at X = -1/3 and 3/2]
call these A and B
call (3,0) C

tan BNC = 3/4
tan ANC = 5/3

With A=BNC, B=ANB, (A+B) =ANC

Using tan (P-Q) = (tan P - tan Q)/ (1+ tanP.tanQ)
[(5/3)-(3/4)]/1+(5/3)(3/4)

tan ANB = tan (ANC - BNC) = 11/27 [QED]

Aitch

[Sorry about the editing - too many As and Bs!]
4. (Original post by dida87)
http://img190.imageshack.us/img190/8396/img17946bt.jpg

sinα = 2/√5
cosα = 1/√5

sinß = 0.6
cosß = 0.8

Stuck on the last part! Any suggestions?
Assuming you mean parts iii and iv:

iii) sin(a + b) = sinacosb + sinbcosa

iv) You could work out cos (a + b) using cosacosb - sinasinb, and then say that tan(a+b) = sin(a+b) ÷ cos(a+b).

Not tried these methods out for myself, but I think they'll work. That's a pretty nice question in my opinion.
5. Cheers people, what was cloudy is now clear!
6. (Original post by dida87)
Cheers people, what was cloudy is now clear!
and he who's rowdy turns to beer.....

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