1200=π(300/cosA)²(0.8)/(0.5π(196)(1+sinA-0.5cosA) after substitution.
Let f(A)=1200-π(300/cosA)²(0.8)/(0.5π(196)(1+sinA-0.5cosA)),
f(A)=1200-36000sec²A/(49(1+sinA-0.5cosA)).
Average slope between two points (A₀,f(A₀)) and (A₁,f(A₁)):
(f(A₁)-f(A₀))/(A₁-A₀). We can abbreviate this as ∆A=∆(f(A))/∆A.
This is a forerunner to Newton’s Method, which computes gradient using calculus. The Secant Method approximates to the calculus.
So the iteration equation is in essence
A=A-f(A)/∆A=A-f(A)∆A/∆(f(A)), where values of A on the right are used to compute a new value of A on the left.
Specifically, for example:
A₂=A₁-f(A₁)(A₁-A₀)/(f(A₁)-f(A₀)).
We are not given any initial values but we could invent some. Because we are dealing with trig functions we can choose some well-known values. Let A₀=0 and A₁=π/6, so sinA₀=0, cosA₀=1, secA₀=1, sinA₁=0.5, cosA₁=√3/2, secA₁=2/√3 or 2√3/3. We can tabulate successive iterations, where r is the iteration number.
Let g(Aᵣ,Aᵣ₋₁)=Aᵣ-f(Aᵣ)(Aᵣ-Aᵣ₋₁)/(f(Aᵣ)-f(Aᵣ₋₁)).
After around 8 or 9 iterations we arrive at A=0.11761 radians (6.7385°) approx.
<Unfortunately a fault in the system prevents me from uploading a table>