Picture Problem -- completely lost!

Trigonometric ratios and functions, the unit circle, inverse trig functions, identities, trig graphs, etc.
Posts: 2
Joined: Tue Apr 16, 2013 6:07 am

Picture Problem -- completely lost!

Postby andrew05 » Tue Apr 16, 2013 6:35 am

I'm having a hard time figuring this problem out, using this image:


The distance AD and BC is 18 feet. The distance between DC is 12 feet. And for clarification, the 8 degrees is between the vertical by AD and the actual AD pole, if that makes any sense.

I need to find the length of the brace AC and the distance AB between the supporting poles at the ground level. If the image is too hard to see, I'll re upload it as a larger document. The bottom left corner is A, bottom right B, top left D, top right C.

I started out trying to find the "invisible" triangle's angles where the 8 degree is one corner, with a right triangle top left, and 82 degrees for the remaining angle up by D. Which if that is true makes the angle ADC 98 degrees. I also got 21.63 feet as the distance AC from the pythagorean theorem; but then I realized the triangle isn't a right triangle so i'm not sure if it can be used. With the law of cosines that side is 22.98 feet so that's more likely. Anyways after this I get weird numbers and get confused; I have no idea if what I did is right and where to go next.

Posts: 100
Joined: Sun Feb 22, 2009 11:12 pm

Re: Picture Problem -- completely lost!

Postby theshadow » Tue Apr 16, 2013 1:09 pm

i'd drop a vertical from D to AB. that makes a right tirangle with angles 82 and 8 and hypotenuse 18. solve the triangle for the height h=18sin82 and b=18cos82. between the verticals is 12. do the triangle with A, C and the base of the vertical from C to AB. this has height h and base b+12. do pythagorean to get AC.

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