The problem states that a semi circular arch over a street that has a radius of 14feet. A banner is attached t

Question

The problem states that a semi circular arch over a street that has a radius of 14feet. A banner is attached to the arch at points A and B, such that AE=EB=5 feet. How many feet above the ground are these points of attachment for the banner?

The problem states that a semi circular arch over a street that has a radius of 14feet. A banner is attached to the arch at points A and B, such that AE=EB=5 feet. How many feet above the ground are these points of attachment for the banner?

Answer 1

I will adopt the assumption that E is the midpoint of the arch. However, the claim that A and B have x-coordinates of +5 and -5 respectively is incorrect; the distance from A (or B) to E is 5 ft, but the points on the circle at x-coordinates of +5 and -5 are also at lower y-values, meaning that these points have a total distance from E that is greater than 5.

If you draw a semicircle and place E at the midpoint of the arc, you can place A to the left, somewhere on the circle. Draw a radius to E and a radius to A. Each has length 14. Draw the chord connecting A and E. We know that the chord has length 5. So now we have two radii with an unknown angle, x, between them, and a leg of known length opposite them. By the law of cosines, we can say 5^2 = 14^2 + 14^2 - 2*24*24*cos(x) ==> 25 = 392(1 - cos(x)) ==> 1 - cos(x) = 0.0638 ==> cos(x) = 0.936 ==> x = 20.57 degrees. Now, the height of point A above the ground is just 14*sin(90 - x), because the radius to A makes the angle 90 - x with the horizontal. 14*sin(90 - 20.57) = 13.1 feet above the ground.

(Note that this answer is very close to the answer with the assumption that the x-coordinates of A and B are +5 and -5, because the actual x-coordinates of A and B turned out to be 14*sin(20.57) = +4.92 and -4.92, but this close only by coincidence. If the distance from the midpoint had been greater, the answer would not have been close at all.)

Answer 2

Hm...
I got the answer 16.9911486

by finding the circumfrence of the whole circle, dividing by 2, subtracting 10 (because the points are both 5 ft away from the middle) and then dividing by 2 again (because the points are on opp. sides). That gave me how far along the arch points A and B were and you were probably looking for how far they are from the ground when two lines are drawn from point A and point B perpendicular to the ground. Thats prob. why I got a different answer than everyone else.

I know I should just erase all that but I'm not going to.

I used law of sines to find the answer:

The hypotenuse is 14 feet and it is a right triangle.
One angle is 5 degrees because of the the definition of arc measure. And that means the last angle in the triangle is 85 degrees. Now you can set up the equation:

sin 85 = sin 90 = sin 5
--------- -------- ----------
x 14 y

now multiply sin 85 by 14, divide by sin of 90 (which is equal to one), to find x. And multiply sin 5 by 14 to find y.

You should get:
x = 1.22
y = 13.9467

y is the line that you are looking for to find the height from the ground, so the answer is:

(apx.) 13.9467

Answer 3

The easy answer:

Assuming E is the mid point of the banner and the banner is straight and horizontal.

From the centre of the arch at ground to the fixing point must be 14 ft (as its a radius).

From the fixing point to the centre of the banner is 5 ft (given)

This is now a right angled triangle with hypotenuse of 14ft and 1given side of 5 ft

From Pythagoras (a^2 = b^2 +c^2)
14^2 - 5^2 = c^2

196 - 25 = 171
SQRT 171 = 13.08

Answer 4

Hard to answer this question unless you can specify further about the point E ??? I'm thinking that E must be the midpoint of the banner??

IF E is the midpoint of the banner, I would proceed as follows:

Draw a semi-circle of radius 14 above the x - axis

Given my midpoint assumption, point A would have

x-coordinate -5 , and point B would have x-coordinate +5

Then use the equation for circle of radius 14 centered at origin

x^2 + y^2 = (14)^2 and solve for y

y^2 = (14)^2 - x^2

y = sqrt [ (14)^2 -x^2 ]

y = sqrt ( 196 - x^2 ) = sqrt ( 196 - 25 ) = sqrt (171)

= 13.08 ft

Answer 5

Wow! I was reading over what the future crop of Nobel Prize winners had so precisely and mathamatically used to come to their final equations. (Kind of like using a 12 Ga. shot gun to swat a fly) I guess.

Answer seems straight forward enough, but I could be wrong. Banner is attached at point "T", which happens to be the interior high point of the 14'-0" radius, (interior dimension), Arch, (where R = 14'-0").

The semil-circle arch is supported by, and attached to the 5'-0" high pair of CMU brick clad pillars on either side of the narrow street. the arch spans, like an inverted "U". The Banner, or ribbon is hung from the center, - interior high point of the Arch , (point "T"), which is 19'-0" above street level, (or the base of the supporting pillars anyway). Then the two equal length Banners / Ribbons remaining, are attached at the inside face of the arch where it meets the inside face of the pillar, - at 5'-" above the base of the pillar, so that the banner "drapes" inside of the Arch.