Expert: Bobby Soltani - 11/27/2005

Question
Two buildings are seperated by an alley. Tom is looking out of a window 60 feet above the ground in one building. He observes the measurement of the angle of depression to the base of the second building to be 50 degrees and the angle of elevation to the top will be 40 degrees. How high is the second building?

Answer
Hi Tiffany,

To solve this problem, we need to first draw a picture.  Draw the first building on the left.  Then, leaving a gap for the alley, draw a second, taller building on the right.  Now draw a straight line from the top corner of the first building to the top corner of the right building.  Draw another line from the top corner of the left building horizontal to the other building.  Finally, draw a line from the top of the first building to the bottom of the second building.  Did you get all that?!

Now, let the distance between the two buildings be x.  Let the height of the first building, which is also the distance between the ground and the horizontal line on the second building, be y1.  Let the difference in height between the two buildings be y2.  Mark your picture accordingly.

You should see two triangles: one with the base x, height y2, and hypotenuse from the top of the left building to the top of the right building.  The other triangle has base x, height y1, and hypotenuse from the tope of the left building to the bottom of the right building.

We know that y1 = 60 feet because it is given.  We can now solve for x using the tangent fuction, since tan(theta) = (opposite)/(adjacent).

The angle of depression is given to be 50 degrees.

tan(50) = 60/x

x = 50.35

Now, we can find y2, because we have the adjacent distance x.

tan(40) = y2/50.35

y2 = 42.25

The height of the second building is y1 + y2, which is 60 + 42.25 = 102.25 feet

Let me know if you have any questions.

Bobby