Story problem from pre-calc.Two buildings of equal height are 800 feet apart. An observer on the street between the buildings measures the angles of elevation to the tops of the buildings as 27...

Story problem from pre-calc.

Two buildings of equal height are 800 feet apart. An observer on the street between the buildings measures the angles of elevation to the tops of the buildings as 27 degrees and 41 degrees, respectively. How high, to the nearest foot, are the buildings?

Asked on by lindsey0

2 Answers | Add Yours

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

Let the nearest distance of the observer from one of the two equally high buildings be x.

The angle of the observing equal height from the nearer point should be higher than that of distant building. So the nearest angle of observation is 41 deg.

Now Let A be the position of the observer, B the feet of the nearest building and C the top most point of the nearest building.

Now ABC is a right angled triangle, with angle ABC=90 deg and angle BAC =41 deg.

Therefore, tan BAC= BC/AB = h/x, where h is the hight.

Therefore, x= h/tan(angleBAC) = h/tan41................(1)

Similarly, the distance to the feet of the other building is 800-x = h/tan27deg............................................(2)

(1)/(2) eliminates h :

x/(800-x) =tan27/tan41 or

x = (800-x)tan27/tan41 or

x(1+tan27/tan41)=800tan27/tan41.

x=800 tan27/(tan41+tan27)=295.6315 feet is the nearest distance of the observer.

The distance between the feet of the other building and the observer is 800-295.6315 feet = 504.3685 feet.

Now, the other unnown is h, the hight of the building.

By equation (1) h=xtan41 = 295.6315*tn41 = 256.9886 feet.

 

 

 

 

krishna-agrawala's profile pic

krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted on

Given:

Both buildings have same height. Let this height be equal to h.

Distance between two buildings = x = 800 feet

Angle of elevation of first building = A1 = 27 degrees.

Angle of elevation of second building = A2 = 41 degrees.

Let us assume that the distance of observer from first building is x1,

and that from the second building is x2.

Then x1 + x2 = Distance between buildings = X = 800 feet.

Tan A1 = h/x1 = Tan 27 = 0.5095

or x1 = h/0.5095

Tan A2 = h/x2 = Tan 41 = 0.8693

or x2 = h/0.8693

But we know: x1 + x2 = 800 feet

Substituting these values of x1 and x2 in above equation we get:

h/0.5095 + h/0.8693 = 800

Multiplying both sides by 0.5095*0.869 we get

0.8693h + 0.5095h = 354.32668

Therefore h = 354.32668/(0.8693 + 0.5095) = 256.9819 Feet

x1 = h/0.5095 = 256.9819/0.5095 = 504 feet

and

x2 = h/0.8693 = 256.9819/0.8693 = 296 feet

Answer:

First building is 504 feet apart and the second building is 296 feet apart.

We’ve answered 318,912 questions. We can answer yours, too.

Ask a question