the angle of elevation of d top of a vertical pole frm a height of 1.54m above a horizontal ground nd d point of observation is 20m frm d pole. calculate d height of d pole to 3.s.f
To solve this problem, we can use trigonometry and the tangent function. The angle of elevation represents the angle between the line of sight from the observer to the top of the pole and the horizontal ground.
Let's denote the height of the pole as "h". According to the problem, the point of observation is 20 m away from the pole, and there is a height difference of 1.54 m between the observer and the ground.
Drawing a right triangle with the horizontal ground as the base, the pole as the vertical side, and the line of sight as the hypotenuse, we can apply the tangent function.
Tangent of the angle of elevation (θ) is equal to the opposite side (height of the pole, h) divided by the adjacent side (distance from the observer to the pole, 20 m).
tan(θ) = h / 20
To find h, we can rearrange the equation:
h = 20 * tan(θ)
Now, we need to calculate the value of tan(θ).
Using a scientific calculator, enter 20, then press the tangent button (tan), and finally, input the angle of elevation, 1.54.
The value of tan(θ) is approximately 0.028945.
Now, we can calculate the height of the pole:
h = 20 * 0.028945
Calculating this expression, h is approximately 0.579 m.
Therefore, the height of the pole is approximately 0.579 m, rounded to 3 significant figures.
i dont know but have a good day bby girl
clearly, 1.54m, as you said.
However, tanθ = 1.54/20
PS
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