Right Triangle Trigonometry

greatwhiteshark

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May 8, 2005
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279
Angle B and C are each 92º. AB and CD each equal 6 feet. The distance from A to D is how much larger than the distance from B to C (to the nearest hundredth).
(You may assume in this figure that the top, the bottom, and side BC are all horizontal. Also assume that a segment connecting point A to point D will be horizontal.)
 
Hello, greatwhiteshark!

Your description doesn't make sense.
First of all, you didn't tell us what we have: a quadrilateral, pentagon, hexagon, . . .

Angle B and C are each 92º. AB and CD each equal 6 feet.
The distance from A to D is how much larger than the distance from B to C (to the nearest hundredth)?

(You may assume in this figure that the top, the bottom, and side BC are all horizontal. ?
Also assume that a segment connecting point A to point D will be horizontal.) ?
With A,B,C,D it seems that we have a four-sided figure.
Then how can the "top, bottom and side" all be horizontal (parallel)?
Then the fourth side, AD, is also horizontal?
I'll take a guess at what the figure looks like . . .
Code:
      B           C 
       *---------*              We have: AB = CD = 6
      /92d     92d\        
    6/             \6        Angle B = Angle C = 92 degrees
    /               \
   *-----------------*
  A                   D
From B drop a perpendicular BE to base AD.
Code:
      B
       *
      /|       Angle ABE = 2 degrees
    6/ |
    /  |             Side AB = 6
   *---*
  A    E
Then: . AE .= .6 sin 2<sup>o</sup> .= .0.20939698 ft

Therefore, AD is about 0.42 feet longer than BC.
 
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