Round off to 5 decimal places

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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>

by Top Rated User (827k points)

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