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 Need Help With Math Problem Related To Triangle.
 J_Hallgren From United States of America, joined Jun 2000, 1507 posts, RR: 0Posted Tue May 31 2005 15:27:24 UTC (10 years 12 months 4 days 19 hours ago) and read 3242 times:

 It's been 30+ yrs since high school for me so... Need to find out how long a side of a triangle would be: Thus starting at point A, I walk straight ahead for 50 feet to point B. I then turn right (90 degrees) and go to unknown point C. How far should I walk to get to C if the angle from point A to C is 20 degrees off from A to B? Does that description make sense? Thanks!
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 Redngold From United States of America, joined Mar 2000, 6907 posts, RR: 40 Reply 1, posted Tue May 31 2005 15:31:38 UTC (10 years 12 months 4 days 19 hours ago) and read 3239 times:

 Here's all you need to know: The Pythagorean Theorem (applies only to right triangles): a^2 + b^2 = c^2 and: All the angles of a triangle must add up to 180 degrees, if you're trying to figure out the degree of angle.
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 MD11Engineer From Germany, joined Oct 2003, 14968 posts, RR: 61 Reply 2, posted Tue May 31 2005 15:35:42 UTC (10 years 12 months 4 days 19 hours ago) and read 3237 times:

 Quickly I would say: Alpha = Angle between AC and AB AB = distance A to B BC = distance B to C BC = tan (Alpha) * AB Jan
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 J_Hallgren From United States of America, joined Jun 2000, 1507 posts, RR: 0 Reply 3, posted Tue May 31 2005 15:38:07 UTC (10 years 12 months 4 days 19 hours ago) and read 3233 times:

 The problem with the application of P-T to this issue is that I only know the length of side A! And I need the length of side B, not the angle of the third corner, which I knew to be 70 degrees. I also don't have any reference material to get the value of tangent for this issue, and search for it turns up way too many items.[Edited 2005-05-31 15:41:38]
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 SLC1 From , joined Dec 1969, posts, RR: Reply 4, posted Tue May 31 2005 15:43:19 UTC (10 years 12 months 4 days 19 hours ago) and read 3228 times:

 if memory serves me correctly, and i understand your question, your length BC=50*sin20/sin70, which ultimately comes to 18.191851171331 this is because sin A/BC = sin B/AC = sin C/AB[Edited 2005-05-31 15:46:30]
 FLYtoEGCC From United Kingdom, joined Feb 2004, 947 posts, RR: 2 Reply 5, posted Tue May 31 2005 15:47:36 UTC (10 years 12 months 4 days 19 hours ago) and read 3222 times:

 You know the angle BAC to be 20 degrees. Tan (angle) = opposite side / adjacent side So if your unknown side BC = x, you would do: Tan 20 = x / 50 So x = 50 * Tan 20 Which should give you an answer of 18.19851171 ft...
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 Redngold From United States of America, joined Mar 2000, 6907 posts, RR: 40 Reply 6, posted Tue May 31 2005 15:48:57 UTC (10 years 12 months 4 days 19 hours ago) and read 3218 times:

 Argggh I didn't read the question well enough... Definitely need the sines and tangents here...
 Up, up and away!
 J_Hallgren From United States of America, joined Jun 2000, 1507 posts, RR: 0 Reply 7, posted Tue May 31 2005 15:51:30 UTC (10 years 12 months 4 days 19 hours ago) and read 3216 times:

 Tried Google with different phrase and found: http://www.pangolin.com/userhelp/scanangles.htm Which deals with scanner and screen but gives the formula and value of Tangent for 20 deg as .364 Edit: I was posting this at same time FlytoEgc did so didn't see that reply but THANKS to all who helped me!  [Edited 2005-05-31 15:58:48]
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