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This is an extra credit question, so I'll pass along 10 points to the person who can show me how they got their answer: "A laser that sits 59" off the floor is aimed at a wall 4 feet away. If the laser is tilted downward by 15 degrees, the dot of light on the wall will be how many inches off the floor?"

I got 53 inches, but the guy who sits beside me got 51 inches. Are either of these correct? Thanks...

2006-08-30 06:01:07 · 6 answers · asked by heaven scent 1 in Education & Reference Homework Help

6 answers

this is what I get:

You have a right triangle, formed by the 4 ft (48 in) distance between the laser and the wall (side a), the unknown distance between the dot of light on the wall and the point on the wall which is 59 inches above the floor (side b),and the length of the beam of light between the laser and the wall (side c, the hypoteneuse of this triangle). Angle B, which is opposite side b, is 15 degrees. Therefore:

tan15 = b/48
b = tan15 x 48
b = 12.86156... (approx 13 inches)

Since 13 inches is the distance between the point on the wall which is 59 inches above the floor and the dot of light, just subtract 13 from 59, which gives you 46 inches.

2006-08-30 06:40:59 · answer #1 · answered by Marcella S 5 · 0 0

h(l) = 59"
l = 4' = 48"

1.) Calculate the sides of the triangle.
Given: Angle (a) = 15 degrees, Adjacent side = 48"
Formula: tan (a) = Opposite/Adjacent
0.2679 = opposite / 48"
12.8592" = opposite

This means the laser light will be 12.8592" lower than the laser's height.

59" - 12.8592" = 46.1408"

2006-08-30 13:09:35 · answer #2 · answered by ³√carthagebrujah 6 · 0 0

x= distance off the floor

where x=59-a

Tan(75)=48/a

a=48/tan(75)

=46.14

2006-08-30 13:08:49 · answer #3 · answered by odu83 7 · 0 0

It's a triangle, and you have side-angle-side. Sorry, but it's been 10 years since I was in geometry, but there's a start.

2006-08-30 13:10:35 · answer #4 · answered by pacerslover31 3 · 1 0

i have no idea

2006-08-30 13:06:20 · answer #5 · answered by 2cute2handle♥ 4 · 0 3

you will never need to use this... sorry....

2006-08-30 13:08:24 · answer #6 · answered by ksgirl 4 · 0 3

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