The angle (less than 90°) that the cable makes with the ground for this considered case is 59° approximately.
Let there is triangle ABC such that |AB| = a units, |AC| = b units, and |BC| = c units and the internal angle A is of θ degrees, then we have:
[tex]a^2 + b^2 -2ab\cos(\theta) = c^2[/tex]
(c is opposite side to angle A)
Consider the diagram attached below.
In the diagram given, we have:
AB = tree with length of 45 ft
AC = cable, with length of 'x' ft (assume)
BC = distance of the base of the cable from the base of the tree of measurement 30 ft.
AD = vertical height of the point A (the top of the tree and cable) from the ground of 'h ft' (assume),
Using the cosine rule for the triangle ABC, taking:
|AC| = x = length of the side opposite to the known angle
|AB| = 45 ft and BC = 30 ft are rest of the two sides,
Angle opposite to AC is of 86°
Thus, we get:
[tex]45^2 + 30^2 -2(30)(45)\cos(86^\circ) = x^2\\\\2925 - 188.34 \approx x^2\\\\x \approx \sqrt{2736.66} \approx 52.31 \: \rm ft[/tex]
(took positive root because of x being length, which is a non-negative quantity).
For triangle ABD, using the sine ratio from the perspective of angle A, we get:
[tex]\sin(m\angle A) = \dfrac{\text{side opposite to A}}{\text{length of hypotenuse}} \\\\\sin(86^\circ) = \dfrac{h}{45}\\\\h = \sin(86^\circ) \times 45 \approx 44.89 \: \rm ft[/tex]
Now, for the triangle ACD, using the sine ratio from the perspective of angle C (smaller than 90°), we get:
[tex]\sin(m\angle C) = \dfrac{\text{side opposite to C}}{\text{length of hypotenuse}} \\\\\sin(\theta^\circ) = \dfrac{h}{x} \approx \dfrac{44.89}{52.31} \approx 0.858\\\\\theta =\sin^{-1}(0.858) \approx 59.09, 120.90[/tex]
Taking the angle smaller than 90°, we get:
[tex]\theta \approx 59.09^\circ \approx 59^\circ[/tex]
Thus, the angle (less than 90°) that the cable makes with the ground for this considered case is 59° approximately.
Learn more about law of cosines here:
https://brainly.com/question/17289163
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