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Question

1.A picture on a wall is 2ft high and 1ft above the eye’s observer. At what distance from the wall should the observer position in order to have a maximum subtended angle on the picture?

2.Approximate sin(3.14159) using differentials correct to 5 decimal places.

Answer 1

1. In the cross-section containing the human's eye and perpendicular to both wall and the floor, let A be the top of the picture, B the bottom of the picture, C below the picture but at the eye's level, and E the eye. Hence AB = 2ft. BC = 1 ft. Our task is to maximize angle AEB, which is bigger than 0 but smaller than pi/2. Let F = tan (angle AEB). When angle AEB is maximized, F is also maximized. Let BEC notes the angle BEC, and AEC notes the angle AEC. Also let h = CE, the distance of the eye from the wall. Hence:
3/h = tan AEC = tan (AEB + BEC)
= (F + tan(BEC)) / (1 - F tan(BEC))
= (F + 1/h)/(1 - F/h)
or 3(h - F) = Fh^2 + h
F = 2h / (h^2 + 3)
dF/dh = 0 when F is at maximum:
0 = dF/dh = { 2(h^2 + 3) - 2h*2h} / (h^2 + 3)^2
leads to h = √3
So the observer position should be at √3 ft from the wall in order to have a maximum subtended angle on the picture.

2. Let y = pi - 3.14159
sin(3.14159) = sin(pi - 3.14159) = sin y
= y - y^3/3! + y^5/5! - ....
≈ y = 0.0000026535....