Maths Querry

[aratti]

The problem im having is i need to use inverse sin ( sin-1) for the following equation:

Inverse sin is not possible with PBP but inverse tan yes (I am sure to have seen an example (ATN2) given by Darell sometime ago).

In this case you can just manipulate a bit your equation to obtain the tangent, since:

SIN = PERP./HYP. (where: (v-v0) = PERP. and (v1-v0) = HYP.)

TAN= PERP./BASE (where: (v-v0) = PERP. and SQR((v1-v0)^2 - (v-v0) ^2) = BASE)

so:

ATN2 ((v-v0)/SQR((v1-v0)^2 - (v-v0) ^2))

should give you your angle (I don't remember if in degrees or in radiants)

Edited: See also this link:http://www.picbasic.co.uk/forum/show...ighlight=angle

Al.