# Trig confusion

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• Jan 14th 2008, 03:17 AM
nugiboy
Trig confusion
Hi

Im a bit confused on working out trig equations. I know all the rules, but something doesn't make sense to me.

There is a question in my excercise book, which i worked out and i ended up with having to solve:

Sin 5 = 117.8 / x

Using the normal rules of equations, i got

x sin 5 = 117.8
117.8 / sin 5 = x
x = 1351

But the book said to do:

117.8 sin 5 = x
x = 10.3

Im a bit confused on why this works. Surely you multiply by the the denominator. Any explanations?
• Jan 14th 2008, 03:26 AM
mr fantastic
Quote:

Originally Posted by nugiboy
Hi

Im a bit confused on working out trig equations. I know all the rules, but something doesn't make sense to me.

There is a question in my excercise book, which i worked out and i ended up with having to solve:

Sin 5 = 117.8 / x Mr F says: Let me guess. You're using trigonometry to find the length of a side in a right-triangle. This expression you got, sin 5 = 117.8 / x, is probably wrong. It's probably meant to be sin 5 = x/117.8 .....

Using the normal rules of equations, i got

x sin 5 = 117.8
117.8 / sin 5 = x
x = 1351

But the book said to do:

117.8 sin 5 = x
x = 10.3

Im a bit confused on why this works. Surely you multiply by the the denominator. Any explanations?

Please give the question in your exercise book - exactly as it is written.
• Jan 14th 2008, 03:43 AM
nugiboy
Quote:

Originally Posted by mr fantastic
Please give the question in your exercise book - exactly as it is written.

Here it is:

http://i137.photobucket.com/albums/q...tiger/trig.jpg

Its part iii im stuck on.

I need to find the force on the horizontal before i can use f = ma to find the acceleration. This basicly simplifies to this:

http://i137.photobucket.com/albums/q220/e3tiger/aha.jpg

(Edit: the angle in this diagram should be 5' not 15')

Sin = opp / hyp

Therefore in this example, wouldnt sin 5 = 117.8 / hyp ?
• Jan 14th 2008, 04:11 AM
mr fantastic
Quote:

Originally Posted by nugiboy
Here it is:

http://i137.photobucket.com/albums/q...tiger/trig.jpg

Its part iii im stuck on.

I need to find the force on the horizontal before i can use f = ma to find the acceleration. This basicly simplifies to this:

http://i137.photobucket.com/albums/q220/e3tiger/aha.jpg

(Edit: the angle in this diagram should be 5' not 15')

Sin = opp / hyp

Therefore in this example, wouldnt sin 5 = 117.8 / hyp ?

No.

Read the attachment. Pay close attention to the triangle showing the resolution of the weight force into components parallel and perpendicular to the slope - note that

$\sin 5^0 = \frac{W_{parallel}}{W} = \frac{W_{parallel}}{117.6} \Rightarrow W_{parallel} = 117.6 \sin 5^0$,

where $W_{parallel}$ is the component of the weight force (W = 117.6) parallel to the slope.