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October 26, 2014

October 26, 2014

Posted by **Nada** on Tuesday, July 20, 2010 at 4:32pm.

- Mathh -
**Henry**, Wednesday, July 21, 2010 at 4:16pmNada, can you send me the answer so I

can compare with mine before posting?

- Mathh -
**Henry**, Thursday, July 22, 2010 at 5:59pmTo help you understand what I've done,

we'll draw a picture of the problem:

1. Draw a rectangle with the longest

sides in hor. positiion and the shortest sides vertically.

2. Draw a diagonal from the upper

left vertex to the lower right vertex.

3. Draw a 2nd line from the upper left

vertex to a point near the center of

the bottom line. Label the portion to

the lt. 270 m and the portion to the rt. X meters.

4. Label the angle between the top

line of the rectangle and the diag-

onal 54 deg. This is the angle of

depression.

5. At the lower right vertex, the diag.

forms 2 angles. Label the one on the left 54 deg. also. We have formed 2 rt. triangles. The angle between the

top line of the rectangle and the shortest hyp. is the 2nd angle

of depression and measures 71 deg.

6. Label the acute angle formed by the smaller hyp. and the bottom of the

rectangle 71 deg.

Tan 71 = h / 270, h = 270 Tan 71 =

784.14 m. = Height of observer.

Tan 54 = 784.14 / (X = 270),

(X + 270) Tan 54 = 784.14

X + 270 = 784.14 / Tan 54 = 569.7

X = 569.7 - 270 = 299.7 m = Length of

Aqueduct

- Mathh -
**Jen**, Tuesday, May 14, 2013 at 11:44amThanks so much !

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