# Rotation of triangle

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• March 30th 2009, 01:25 PM
Ian1779
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Rotation of triangle
Hi

I am finding this question tough (attached), I think I may have stumbled on the answer but could do with some confirmation I'm on the right path.

Right, so far I have found the length $BC = \sqrt 82$ using Phythagoras.

Now I need to determine $sin \theta$, $cos \theta$ and $tan \theta$

This is where I am stuck. It is not a right angled triangle so can I use SOHCAHTOA?
Never the less I have attempted this way and got

$tan \theta = 1/9$

$sin \theta = 1/\sqrt82$

$cos \theta = 9/\sqrt82$

Which all give the same answer incidentally - is this the right approach or am I off in the completely wrong direction?

I get the last part about the two line notation, but am not sure what values to use for x and y.

Thanks
• March 31st 2009, 12:18 PM
earboth
1 Attachment(s)
Quote:

Originally Posted by Ian1779
Hi

I am finding this question tough (attached), I think I may have stumbled on the answer but could do with some confirmation I'm on the right path.

Right, so far I have found the length $BC = \sqrt 82$ using Phythagoras.

Now I need to determine $sin \theta$, $cos \theta$ and $tan \theta$

This is where I am stuck. It is not a right angled triangle so can I use SOHCAHTOA?
Never the less I have attempted this way and got

$tan \theta = 1/9$

$sin \theta = 1/\sqrt82$

$cos \theta = 9/\sqrt82$

Which all give the same answer incidentally - is this the right approach or am I off in the completely wrong direction?

I get the last part about the two line notation, but am not sure what values to use for x and y.

Thanks

After the rotation the side B''C'' has the slope m = 0. Calculate the the slope of BC (2-point-formula of a straight line) which is equal of the tangens of the angle of rotation:

$m=\dfrac{2-1}{-4-5}=-\dfrac19$

That means you have to rotate the triangle by an angles whose tangens is $+\dfrac19$

I'm not quite sure what is meant by the two-line notation. The angle in question is included by the lines with the equations:

$l_1: y = 0\text{ and }l_2:y = -\dfrac19 \cdot x$
• March 31st 2009, 01:20 PM
Ian1779
Thanks so much. The drawing helped even more so I could visualise it better.
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