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Two spotlights, one blue and the other white, are placed 6 m apart on a rack in the ceiling of a ballroom. A stationary observer standing on the ballroom floor noticed the angle of elevation is 45 degrees to the blue spotlight and 70 to the white. How high, to the nearest tenth of a meter is the ceiling of the ballroom.

 

My answer is 4m off according to answer in the textbook and is way above margin of error so I know I'm missing something and anyone who can show how to properly solve this would be greatly appreciated.

Guest Jan 21, 2016
edited by Guest  Jan 21, 2016
edited by Guest  Jan 21, 2016
edited by Guest  Jan 21, 2016

Best Answer 

 #1
avatar+78618 
+10

We can solve this as follows :

 

Let the height of the ceiling  = H

 

So, from the observer's position........we have

 

tan (45) = x/H        and      tan (20)  = (6 -x)/H        

 

Where x  and 6-x are the horizontal distances from the observer to lines drawn from each light perpendicular to the floor [ i.e, the ceiling height under both lights]

 

Solving for H, we have

 

x / tan(45)  =  (6-x)/tan(20)         cross-mutiply

 

xtan(20 )= (6-x)tan(45)   simplify

 

xtan(20) = 6tan(45) - xtan(45)

 

xtan(20) + xtan(45)  = 6tan(45)

 

x[ tan(20) + tan(45)]  = 6tan(45)

 

x = 6tan(45) / [ tan(20) + tan(45)]  =  about 4..399 m

 

So....the height of the ceiling  is given by 4..399/tan(45)  = about  4.399 m

 

P.S.   ....check that   (6 - 4.399)/ tan(20)   also gives you, roughly, 4.399m

 

Here's a pic .....

 

 

The observer is at A and the lights are at D and C  .... AB is the ceiling height =   about 4.399 ...  [4.4 rounded]

 

 

cool cool cool

CPhill  Jan 21, 2016
Sort: 

3+0 Answers

 #1
avatar+78618 
+10
Best Answer

We can solve this as follows :

 

Let the height of the ceiling  = H

 

So, from the observer's position........we have

 

tan (45) = x/H        and      tan (20)  = (6 -x)/H        

 

Where x  and 6-x are the horizontal distances from the observer to lines drawn from each light perpendicular to the floor [ i.e, the ceiling height under both lights]

 

Solving for H, we have

 

x / tan(45)  =  (6-x)/tan(20)         cross-mutiply

 

xtan(20 )= (6-x)tan(45)   simplify

 

xtan(20) = 6tan(45) - xtan(45)

 

xtan(20) + xtan(45)  = 6tan(45)

 

x[ tan(20) + tan(45)]  = 6tan(45)

 

x = 6tan(45) / [ tan(20) + tan(45)]  =  about 4..399 m

 

So....the height of the ceiling  is given by 4..399/tan(45)  = about  4.399 m

 

P.S.   ....check that   (6 - 4.399)/ tan(20)   also gives you, roughly, 4.399m

 

Here's a pic .....

 

 

The observer is at A and the lights are at D and C  .... AB is the ceiling height =   about 4.399 ...  [4.4 rounded]

 

 

cool cool cool

CPhill  Jan 21, 2016
 #2
avatar+91001 
+5

Nice work Chris :))

Melody  Jan 21, 2016
 #3
avatar+78618 
+5

Thanks, Melody....!!!

 

 

cool cool cool

CPhill  Jan 21, 2016

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