1. ## 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 $\displaystyle BC = \sqrt 82$ using Phythagoras.

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

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

$\displaystyle tan \theta = 1/9$

$\displaystyle sin \theta = 1/\sqrt82$

$\displaystyle cos \theta = 9/\sqrt82$

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

2. 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 $\displaystyle BC = \sqrt 82$ using Phythagoras.

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

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

$\displaystyle tan \theta = 1/9$

$\displaystyle sin \theta = 1/\sqrt82$

$\displaystyle cos \theta = 9/\sqrt82$

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:

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

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

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:

$\displaystyle l_1: y = 0\text{ and }l_2:y = -\dfrac19 \cdot x$

3. Thanks so much. The drawing helped even more so I could visualise it better.