refrigerator

[ocngirlbl]

A refrigerator is tipped against a vertical wall so it can be serviced. It makes an angle of 70 degrees with the horizontal floor. How high is the top right corner of the refrigerator above the floor?

 

[galactus]

That depends on how tall the fridge is.

 

[ocngirlbl]

Oh sorry! I forgot to add that information! The refrigerator is 2 meters tall and 1 meter wide

 

[BigGlenntheHeavy]

\(\displaystyle If, \ the \ fridge \ is \ in \ the \ shape \ of \ a \ rectangular \ parallelpiped, \ then \ we \ have:\)

\(\displaystyle sin(20^0) \ = \ \frac{x}{2} \ \implies \ x \ \dot= \ .684.\)

\(\displaystyle and \ cos(20^0) \ = \ \frac{y}{1} \ \implies \ y \ \dot= \ .940.\)

\(\displaystyle Hence, \ x \ + \ y \ \dot= \ 1.624m, \ the \ distance \ of \ the \ top \ right \ corner \ of \ the \ fridge\)

\(\displaystyle from \ the \ floor.\)

 

[ocngirlbl]

Thank you! But where did you get the 20 degrees from to use cosine and sine?

 

[BigGlenntheHeavy]

\(\displaystyle 70^0 \ + \ corner \ of \ fridge(90^0)+20^0 \ = \ 180^0, \ angle \ of \ floor.\)

\(\displaystyle See \ graph \ below. \ Note: \ Not \ drawn \ to \ scale.\)

[attachment=0:37k9v2k6]bbb.JPG[/attachment:37k9v2k6]

 

[ocngirlbl]

Oh, alright! Thanks again!

 

[ocngirlbl]

thank you for posting the picture! on the diagram I got with the problem, the refrigerator was actually leaning left against the wall rather than right, and then angle of 70 degrees was in the corner between the fridge and the wall instead of outside. does this change my result/the way I should approach the problem?

 

[Deleted member 4993]

thank you for posting the picture! on the diagram I got with the problem, the refrigerator was actually leaning left against the wall rather than right, and then angle of 70 degrees was in the corner between the fridge and the wall instead of outside. does this change my result/the way I should approach the problem?

If you understand the solution for the problem above - you should see that the approach will remain same but the answer may change.

 

[ocngirlbl]

so then using the approach from above, my new formulas would be sin(70) = x/2 and cos(70) = y/1 which would be 1.88 + 0.34 = 2.22 meters above the floor. Any objections?

 

[Deleted member 4993]

so then using the approach from above, my new formulas would be sin(70) = x/2 and cos(70) = y/1 which would be 1.88 + 0.34 = 2.22 meters above the floor. Any objections?

Refering to BGTH's drawing - the angle between the house-wall and the refrigerator(back wall) is 70° and the angle between the house-floor and the refrigerator (bottom) is also 70°. The angle between the refrigerator back-wall and the house-floor was 20°.

The calculation you made are for:

the angle between the house-wall and the refrigerator(back wall) is 20° and the angle between the house-floor and the refrigerator (bottom) is also 20°. The angle between the refrigerator back-wall and the house-floor is 70°.