2nd Level Vectors question. Tan ratios.

[Darth Frodo]

If I am only ginen the following how can I solve the whole Tan ratio part?

If vector U is 10 units

"Alpha Symbol" = Tan -1 (Tan inverse) 3/4 find U Vectors V = 13 units

"Beta" = Tan -1 (Tan inverse) 3/4 find V
I tried to multiply the 2 Tan ratios together but no luck.

I imagine you must work both V & U out individually.

I can add vectors and get their i & j components but this ratio has me stumped.

Thanks.

DF

 

[Mark44]

If I am only ginen the following how can I solve the whole Tan ratio part?

If vector U is 10 units

"Alpha Symbol" = Tan -1 (Tan inverse) 3/4 find U


Vectors V = 13 units

"Beta" = Tan -1 (Tan inverse) 3/4 find V



I tried to multiply the 2 Tan ratios together but no luck.

I imagine you must work both V & U out individually.

I can add vectors and get their i & j components but this ratio has me stumped.

Thanks.

DF

How are [itex]\alpha[/itex] and [itex]\beta[/itex] related to the vectors U and V? Are they the angles between the vectors and the positive x-axis?

 

[Darth Frodo]

First, thanks for the response.

"Alpha" is the angle between Vector U and the negative side of X axis

"Beta" is the angle between Vector V and the positive side of X axisNote: Both vectors stem from origin

 

[Mark44]

V = <10cosβ, 10sinβ>
U can be written similarly, but you will need to find the angle that U makes with the positive x-axis.

Since β = tan^-1(3/4), then tanβ = 3/4. Draw a right triangle whose legs are 3 and 4, and you should be able to find sinβ and cosβ.

You can do something similar to find the sine and cosine of the other angle (the angle that U makes with the pos. x-axis).