The lower edge of a 4 foot tall painting is 1 ft. above you eye level.

Question

The lower edge of a 4 foot tall painting is 1 ft. above you eye level.

At what distance should you stand from the wall so you stand from the wall so your viewing angle of the painting is maximized?

_________ft.

Give you answer accurate to at least 2 decimal places.

Thank you in advance.

Answer 1

Question:

The lower edge of a 4 foot tall painting is 1 ft. above you eye level.

At what distance should you stand from the wall so you stand from the wall so your viewing angle of the painting is maximized?

_________ft.

Give you answer accurate to at least 2 decimal places.

Solution:

Draw diagram

. /--------------| 5' <-- Top of Picture Frame

. h2 /----------/ | 4'

. /---------/ | 3' Wall

. /---------/ h1 | 2'

. /--------/ Theta /------------------------| 1' <-- Bottom of Picture Frame

/================================== Eye Level

. x

h1 - hypotenuse of lower triangle with height = 1

h2 - hypotenuse of upper triangle with height = 5

Theta is the angle between h1 and h2

From the graph we can see:

h1 = SQRT(x^2 + 1^2) = SQRT(x^2 + 1)

h2 = SQRT(x^2 + 5^2) = SQRT(x^2 + 25)

The side opposite theta is 4.

Using the Law of Cosines,

c^2 = a^2 + b^2 - 2ab Cos(C)

Solving for C gives:

2ab Cos(C) = a^2 + b^2 - c^2

Cos(C) = [a^2 + b^2 - c^2] / 2ab

C = arccos ([a^2 + b^2 - c^2] / [2ab])

Substitute::

a = h1

b = h2

c = 4

C = Theta

Theta = arccos ([[SQRT(x^2 + 1)]^2 + [SQRT(x^2 + 25)]^2 - 4^2] / [2 * SQRT(x^2 + 1) * SQRT(x^2 + 25)])

Theta = arccos ([x^2 + 1 + x^2 + 25 - 16] / [2 * SQRT(x^2 + 1) * SQRT(x^2 + 25)])

Theta = arccos ([2*x^2 + 10] / [2 * SQRT(x^2 + 1) * SQRT(x^2 + 25)])

To find the maximum angle, take the derivative of Theta and set to 0.

d Theta / dx = d arccos(u) / dx = -u' / SQRT(1-u^2)

d Theta / dx = [4x(x^2 - 5)] / [a lot of other stuff per Wolfram Alpha] = 0

We only need the numerator when we set it equal to 0.

4x(x^2 - 5) = 0

x = 0, +SQRT(5), -SQRT(5)

Throw out 0 since the viewing angle is terrible.

Throw out -SQRT(5) since we cannot go behind the picture.

Therefore, x = SQRT(5)

SQRT(5) = 2.236

THEREFORE, the maximum viewing angle is 2.24 feet away from the wall.