Recent content by mamort

Thanks for monitoring this problem, I have now found the solution which is achieved by solving simultaneous equations of both straight line functions (m1 and m2) that are perpendicular in the top diagram above. The bottom diagram is NOT how the solution is achieved.

Sorry I didn't read the rest, arcos(x) is the inverse of cos(x) not the derivative. Derivative is as follows d/dx cos(x) = -sin(x) d/dx sin(x) = cos(x) This is because sin(x) and cos(x) are essentially the same function except the phase between the two is pi/2 (90 degrees or 1/4 revolution of...

I think it is taken in context of a trig function in which cosine is described by a circle with a radius equal to one. Then since the cos and it's inverse cancel each other out leave the answer is as follows cos(cos-1(x)) = x This would simply mean that the line would run along the x axis, in...

Can someone please look at the diagram below and tell me how u1 is obtained. If it is through the use of m3 please explain how the gradient m3 is obtained.

True I think I might of head down the wrong path I am trying to solve the following problem in the diagram below. I am unsure how u1 is obtained. I assume that the gradient of the green line is used to solve as shown below m3 = (hz+delta_hz)/(u1) therefore u1 = (hz+delta_hz)/m3 where m3 =...

Please take a look at attached diagram displaying gradients of each line, please explain how m3 is obtained.