a vertical pole 3 meters tall casts a shadow 1.8 meters long. what is the angle of the elevation of rhe sun to the nearest minute?
Did you make a diagram?
let the angle be Ø, then 3 is opposite, and 1.8 is adjacent
tanØ = 3/1.8 = 1.666667
Ø = 59.0°
To find the angle of elevation of the sun, you can use trigonometry and the concept of similar triangles.
The vertical pole and its shadow form two right-angle triangles that are similar. In other words, the ratios of their corresponding sides are equal.
Let's denote the height of the vertical pole as h and the length of its shadow as s. In this case, h = 3 meters and s = 1.8 meters.
Now, let's define the angle of elevation of the sun as θ. We need to find the measurement of θ.
Using the concept of similar triangles, we can set up the following equation:
tan(θ) = h / s
Plugging in the values we have:
tan(θ) = 3 / 1.8
Now, we need to find the value of θ. To do this, we need to take the inverse tangent (arctan) of both sides of the equation:
θ = arctan(3 / 1.8)
Using a calculator, we find:
θ ≈ 57.99 degrees
However, the question asks for the angle of elevation to the nearest minute. To convert degrees to minutes, remember that there are 60 minutes in one degree.
So, multiplying 57.99 degrees by 60:
θ ≈ 3479.4 minutes
Rounding to the nearest minute, the angle of elevation of the sun is approximately 3479 minutes.