Asource of laser light sends rays ab and ac toward two opposite walls of a hall. the light rays strike the walls at points b and c, as shown below: a source of laser light is at point a on the ground between two parallel walls. the walls are perpendicular to the ground. ab is a ray of light which strikes the wall on the left at point b. the length of ab is 60m. ac is a ray of light which strikes the wall on the right at point c which is 40m above the ground. the ray ab makes an angle of 60 degrees with the ground. the ray ac makes an angle of 45 degrees with the ground. what is the distance between the walls? 80 m 140 m 70 m 110 m

70 m

Step-by-step explanation:

AB makes a 30-60-90 triangle with the wall.

AC makes a 45-45-90 triangle with the wall.

In a 30-60-90 triangle, the short leg (opposite of the 30° angle) is half the hypotenuse.

In a 45-45-90 triangle, the legs are the same length.

So the distance between the left wall and the laser is 60/2 = 30 m.

The distance between the right wall and the laser is 40 m.

The total distance is therefore 70 m.

  5.   y=6; de=7; fg=8

  6.   x=4.2; jl=14.2; mn=13

step-by-step explanation:

when chords cross inside a circle, the product of the segment lengths is a constant.

5. dh×eh = fh×gh

  y = gh = dh×eh/fh = 3×4/2 . . . divide by the coefficient of gh

  y = 6

of course, the total chord length is the sum of the lengths of its segments.

  de = dh +eh = 3+4 = 7

  fg = fh +gh = 2+6 = 8

6. this problem works the same way as the previous one.

  x = lk = mk×nk/jk = 6×7/10 = 4.2

  jl = jk +lk = 10 +4.2 = 14.2

  mn = mk +nk = 6 +7 = 13

comment on chords and secants

you can think of the points h (problem 5) and k (problem 6) as points where the chords meet. a chord is part of a secant line, a line that intersects the circle in two points. one can also consider these points (h or k) to be the points where the respective secant lines meet.

then, the product that is a constant is the product of the distance from the meeting point (h or k) to one circle intersection with the distance from that meeting point to the other circle intersection. for problem 5, the constant is the product hd×he = hf×hg.

it turns out that this rule regarding the products of lengths to points of intersection is also true when the secants intersect outside the circle.

that is, you only need to remember one rule to work both kinds of problems.

then, the surface area of the sphere shown in the image is: 256π sq. units!

step-by-step explanation:

the lateral surface area of a sphere is 2πrh where h=2r.

then:

surface area = 4πr^2.

in this case, r=8 units ⇒ surface area = 4π(8 units)^2. = 256π sq. units!

then, the surface area of the sphere shown in the image is: 256π sq. units!