# Statue

On the pedestal high 4 m is statue 2.7 m high. At what distance from the statue must observer stand to see it in maximum viewing angle? Distance from the eye of the observer from the ground is 1.7 m.

Result

x =  3.39 m

#### Solution:

$\alpha(x) = \arctan \dfrac{ 4+2.7-1.7}{x}- \arctan \dfrac{ 4-1.7}{x} \ \\ \alpha(x) = \arctan \dfrac{ 5}{x}- \arctan \dfrac{ 2.3}{x} \ \\ \ \\ \alpha'(x) = - \dfrac{ 5 } {x^2+ 25} + \dfrac{ 2.3 } {x^2+ 5.29} \ \\ \ \\ \alpha'(x) = 0 \ \\ \ \\ \dfrac{ 5 } {x^2+ 25} = \dfrac{ 2.3 } {x^2+ 5.29} \ \\ \ \\ x = \sqrt{ 5 \cdot 2.3 } = 3.39 \ \text{m}$

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