explain with solution
Answers
Answer:
1. Find the central angle of the circular sector.
The length of the circular sector is the circumference of the cone’s base, which is 2πr = 2π(20) = 40π, where r is the radius of the cone’s base.
The circular sector’s radius is R = 60, the slant height of the cone. We can also express the arc length as its radius times the central angle, which is 60(θ). So we have:
40π = 60(θ)
θ = (40π)/60 = 2π/3
2. Place points A and B on the circular sector.
Point A can be placed on anywhere on the arc, as those points correspond to the cone’s base. It is helpful to place A on a corner, so that we can place point B on the opposite radius of the arc to correspond to a track around the cone. Point B is 10 units from the base and 50 units from the vertex, so we place it accordingly on the circular sector.
Now we can find the shortest path between A and B quite simply: it is the straight line between the two points! This creates a triangle with segments from A to the vertex, B to the vertex, and AB.
3. Solve for the length of AB.
We can now solve for AB using the law of cosines, as we have a triangle with sides 50, 60, and an angle in between of (2π)/3.
AB2 = 602 + 502 – 2(60)(50) cos (2π/3)
AB2 = 9100
AB = 10√(91)
4. Identify the downhill portion of AB.
It is given in the problem the line AB first goes uphill (increasing distance from the base/decreasing distance from the vertex), and then downhill (decreasing distance from the base/increasing distance from the vertex).
(If you want to verify this is true, you can see my Desmos graphing page where the derivative of the distance shows the track follows exactly this pattern).
Between the uphill and downhill portion is a single point on AB that is closest to the cone’s vertex. The line between this point and the cone’s vertex will be perpendicular to AB, so we form two right triangles for the downhill and uphill portions.
5. Solve for the downhill track length.
Let x be the downhill length, so then AB – x = 10√(91) – x is the uphill length. Let h be the distance from the cone’s vertex to AB. Now we can use the Pythagorean Theorem for the two right triangles to get:
(10√(91) – x)2 + h2 = 602
x2 + h2 = 502
Now here’s a neat trick: subtract the second equation from the first. Then the x2 terms and h2 terms will cancel. So we get:
9100 – 2(10√(91))x = 602 – 502
2(10√(91))x = 8000
x = 400/√(91)
So that’s the answer! This was the 4th answer choice of 400/√(91).
What a remarkable question this was! I could not imagine solving this in the time constraints of a test. But it was fun to solve, and the problem depends on so many different mathematical concepts.......
Hope my this Answer will help u
Answer:
d
Step-by-step explanation:
a photo of sokution enclosed. opening the Cone nappe and plotting the straight line is the trick here.