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Sagot :
[tex]tg(\theta)=\frac{8}{12}\\
\\
tg(\theta)=0.666... \rightarrow \theta\approx35^o[/tex]
The angle of elevation at which he kicks the ball is; 36.89°
Finding the angle of elevation
We are told that the ball is 12 ft away and so we will consider it to be the horizontal distance.
The post is 8 ft high and so we will consider it to be the vertical distance.
Thus, we can use trigonometric ratio to find the angle of elevation. Thus;
tan θ = 8/12
θ = tan⁻¹0.75
θ = 36.89°
In conclusion the angle of elevation at which he kicks the ball is 36.89°
Read more about Angle of elevation at; https://brainly.com/question/19594654
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