Round off to 5 decimal places

in Other Math Topics by

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

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)),


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:


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)

Related questions

1 answer
asked Feb 18 in Other Math Topics by anonymous | 42 views
1 answer
asked Feb 18 in Other Math Topics by anonymous | 28 views
Welcome to, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
86,162 questions
92,163 answers
23,903 users