Recent content by alexm

Yes that's right, which should mean that the cosine rule based answer is correct? Out of interest is there a more elegant solution I may have missed?

From triangle CFE: LEF = LCEtan(θ) LCF = LCE / cos(θ) considering lengths: LBE = LBF + LEF so: LBF = LBE - LEF LBF = LBE - LCEtan(θ) also: LAC = LAF + LCF LAF = LAC - LCE / cos(θ) Now if we say: β = ∠AFB, using the law of cosines gives: d2 = LBF2 + LAF2 - 2LBFLAFcos(β) Which is nearly there...

Apologies they were in the first picture, will change the description.

Edited the original post, must've been trying to do so as it was moved between forums earlier.

Hello @BvU Thanks for your reply :smile:. It's for university homework if that counts? Yesterday I got as far as showing that sin2θ + cos2θ = 1 through my trig efforts, so any help would be much appreciated!

Summary:: We have a rotating arm, offset from the centre of rotation by a certain length, which is controlled by varying the length of a control rod. Need the angle of the rotating arm in terms of length of the rod. The blue line is a fixed column structure. CE and BD form the rotational...