Answer:
(i) [tex]OE = 0.6\cdot \sin \theta + 1.4\cdot \cos \theta[/tex], (ii) [tex]\theta \approx 33.368^{\circ}[/tex], (iii) The maximum value of OE is approximately 1.523 meters, which is associated with an angle of approximately 23.199º.
Step-by-step explanation:
(i) From Geometry, we get that sum of internal angles of trangle AOD.
[tex]\angle A + \angle O + \angle D = 180^{\circ}[/tex] (1)
If we know that [tex]\angle O = 90^{\circ}[/tex] and [tex]\angle A = \theta[/tex], then the value of [tex]\angle D[/tex] is:
[tex]\angle D = 180^{\circ}-90^{\circ}-\theta[/tex]
[tex]\angle D = 90^{\circ}-\theta[/tex] (2)
But we also have the following identity:
[tex]\angle D +\angle D' +\angle D'' = 180^{\circ}[/tex] (3)
If we know that [tex]\angle D = 90^{\circ}-\theta[/tex] and [tex]\angle D' = 90^{\circ}[/tex], then the value of [tex]\angle D''[/tex] is:
[tex]90^{\circ}-\theta +90^{\circ}+\angle D'' = 180^{\circ}[/tex]
[tex]\angle D'' = \theta[/tex] (4)
By Trigonometry, we derive the following formula:
[tex]OE = OD +DE[/tex]
[tex]OE = AD\cdot \sin A +CD\cdot \cos D''[/tex] (5)
If we know that [tex]AD = 0.6\,m[/tex], [tex]CD = 1.4\,m[/tex], [tex]A = \theta[/tex] and [tex]D'' = \theta[/tex], then the value of OE is:
[tex]OE = 0.6\sin \theta + 1.4\cos \theta[/tex] (6)
(ii) If we know that [tex]OE = 1.1\,m[/tex], then the value of [tex]\theta[/tex]:
[tex]0.6\cdot \sin \theta + 1.4\cdot \cos \theta = 1.1[/tex]
By trial and error, we find that [tex]\theta \approx 33.368^{\circ}[/tex].
(iii) Let [tex]OE = 0.6\cdot \sin \theta + 1.4\cdot \cos \theta[/tex], the first and second derivatives of the function are, respectively:
[tex]OE' = 0.6\cdot \cos \theta -1.4\cdot \sin \theta[/tex] (7)
[tex]OE'' = -0.6\cdot \sin \theta -1.4\cdot \cos \theta[/tex] (8)
We equalize the first derivative of the function to zero and solve for [tex]\theta[/tex]:
[tex]0.6\cdot \cos \theta - 1.4\cdot \sin \theta = 0[/tex]
[tex]1.4\cdot \sin \theta = 0.6\cdot \cos \theta[/tex]
[tex]\tan \theta = \frac{0.6}{1.4}[/tex]
[tex]\theta \approx \tan^{-1} \frac{0.6}{1.4}[/tex]
[tex]\theta \approx 23.199^{\circ}[/tex]
And we evaluate the second derivative:
[tex]OE'' = -0.6\cdot \sin 23.199^{\circ}-1.4\cdot \cos 23.199^{\circ }[/tex]
[tex]OE'' = -1.523[/tex]
Then, the critical value is associated with an absolute maximum.
The maximum value of OE is:
[tex]OE = 0.6\cdot \sin 23.199^{\circ}+1.4\cdot \cos 23.199^{\circ }[/tex]
[tex]OE \approx 1.523\,m[/tex]
The maximum value of OE is approximately 1.523 meters, which is associated with an angle of approximately 23.199º.
