i understand the theory i.e i have to prove it is linear interpolation however i can't seem to obtain the rearranged version for r - mainly the c-d part why is that needed and not just d for the similar triangles?, could someone walk it through please.
Also, on q3 am i right in thinking i have too minus the (pi divided by 2) then divide by -3?
As I said I used the 2 small triangles and then had to do manipulation to get the format required
But ... as my diagram shows it was very easy
I did the 1st part in this way,
The gradient of the line(using points P(a,c) and Q(b,d) ) is, m=b−ad−c
Now form the equation of the line, using the point R(r,0), y−0=b−ad−c(x−r)⟹y(b−a)=(d−c)(x−r)⟹y(b−a)=dx−dr−cx+cr⟹y(b−a)=x(d−c)+r(c−d)⟹y(b−a)=−x(c−d)+r(c−d)⟹y(c−db−a)=−x+r⟹r=x+y(c−db−a)
I did similar to this before I realised the easy way I have shown above
That is what I meant when I said it took too much manipulation
Your way is better, but it is difficult to realize that technique. Similar triangles aren't much used at A-Level, so the students will probably not spot the easy technique.
Though my way is also not very long, it only takes around a minute.
Your way is better, but it is difficult to realize that technique. Similar triangles aren't much used at A-Level, so the students will probably not spot the easy technique.
Though my way is also not very long, it only takes around a minute.
Yeah ... from experience I just knew it should be "see-able" rather than "workable"
Your way is better, but it is difficult to realize that technique. Similar triangles aren't much used at A-Level, so the students will probably not spot the easy technique.
Though my way is also not very long, it only takes around a minute.
Similar triangles is widely used as a method for linear-interpolation.