Trent is skiing on a circular ski trail that has a radius of 0.9 km. Trent starts at the 3-o'clock position and travels 2.6 km in the counter-clockwise direction.
How many radians does Trent sweep out?
____ radians
When Trent stops skiing, how many km is Trent to the right of the center of the ski trail?
____ km
When Trent stops skiing, how many km is Trent above of the center of the ski trail?
_____ km
Well, Trent sure knows how to make a roundabout adventure! Let's solve these questions for him.
To figure out how many radians Trent sweeps out, we can use the formula: arc length = radius * angle (in radians). Trent travels 2.6 km, and since the radius is given as 0.9 km, we can plug these values into the formula:
2.6 km = 0.9 km * angle (in radians)
Now, let's solve for the angle:
angle (in radians) = 2.6 km / 0.9 km = 2.888...
Trent sweeps out approximately 2.89 radians. Now, for the next question!
To find how many km Trent is to the right of the center of the ski trail, we need to calculate the horizontal distance he travels. Since he starts at the 3-o'clock position and moves counter-clockwise, he ends up on the left side of the center. The total circumference of the circle is 2 * π * radius, which is 2 * π * 0.9 km = 5.65 km. So, Trent is 2.6 km to the right (or rather, the left) of the center.
Lastly, let's determine how far Trent is above the center. Since the radius is 0.9 km, he is always 0.9 km above the center of the ski trail.
So, Trent is 2.89 radians, 2.6 km to the right (or left) of the center, and 0.9 km above it. Hope Trent had a blast skiing in circles!
To find the number of radians Trent sweeps out, we can use the formula:
radians = distance / radius
In this case, the distance traveled by Trent is 2.6 km, and the radius of the ski trail is 0.9 km.
Substituting these values into the formula:
radians = 2.6 km / 0.9 km = 2.888888888888889 radians
Therefore, Trent sweeps out approximately 2.89 radians.
To find how far to the right of the center Trent is when he stops skiing, we can calculate the horizontal displacement. To do so, we can use trigonometry. The horizontal displacement is given by:
horizontal displacement = radius * sin(angle)
Since Trent is at the 3-o'clock position, which is 0 degrees, the angle is 0 radians. Substituting this into the formula:
horizontal displacement = 0.9 km * sin(0 radians) = 0 km
Therefore, Trent is 0 km to the right of the center when he stops skiing.
To find how far above the center Trent is when he stops skiing, we can calculate the vertical displacement. To do so, we can again use trigonometry. The vertical displacement is given by:
vertical displacement = radius * cos(angle)
Since Trent is at the 3-o'clock position, which is 0 degrees, the angle is 0 radians. Substituting this into the formula:
vertical displacement = 0.9 km * cos(0 radians) = 0.9 km
Therefore, Trent is 0.9 km above the center when he stops skiing.
To find the number of radians Trent sweeps out, we need to use the formula:
θ = s / r
where θ is the angle in radians, s is the arc length, and r is the radius of the circle. In this case, Trent travels an arc length of 2.6 km and the radius is 0.9 km.
θ = 2.6 km / 0.9 km
θ ≈ 2.89 radians
So, Trent sweeps out approximately 2.89 radians.
To find how many kilometers to the right of the center Trent is when he stops skiing, we need to use trigonometry. We can consider a right triangle formed by the radius of the ski trail and the horizontal distance from the center to Trent's position when he stops.
Using the Pythagorean theorem, we have:
x^2 = (0.9 km)^2 - (2.6 km - 0.9 km)^2
x ≈ √(0.9 km)^2 - (2.6 km - 0.9 km)^2
x ≈ √(0.9 km)^2 - (1.7 km)^2
x ≈ √0.81 km^2 - 2.89 km^2
x ≈ √-2.08 km^2 ≈ undefined
The result is undefined because the calculation gave us a negative value under the square root. This means that Trent's position is not to the right of the center of the ski trail.
To find how many kilometers above the center Trent is when he stops skiing, we can use the same right triangle as before.
Using the Pythagorean theorem, we have:
y^2 = (2.6 km - 0.9 km)^2 - (0.9 km)^2
y ≈ √(2.6 km - 0.9 km)^2 - (0.9 km)^2
y ≈ √(1.7 km)^2 - (0.9 km)^2
y ≈ √2.89 km^2 - 0.81 km^2
y ≈ √2.08 km^2 ≈ 1.44 km
Therefore, when Trent stops skiing, he is approximately 1.44 km above the center of the ski trail.
s = rθ, so
(a) θ = 2.6/0.9 = ___
Now think of the unit circle and its relation to the trig functions.
(b) 0.9 cosθ
(c) 0.9 sinθ