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Need Help With Math Problem Related To Triangle.  
User currently offlineJ_Hallgren From United States of America, joined Jun 2000, 1507 posts, RR: 0
Posted (9 years 4 months 3 weeks 3 days 20 hours ago) and read 2132 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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7 replies: All unread, jump to last
 
User currently offlineRedngold From United States of America, joined Mar 2000, 6907 posts, RR: 44
Reply 1, posted (9 years 4 months 3 weeks 3 days 20 hours ago) and read 2129 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.



Up, up and away!
User currently onlineMD11Engineer From Azerbaijan, joined Oct 2003, 14060 posts, RR: 62
Reply 2, posted (9 years 4 months 3 weeks 3 days 20 hours ago) and read 2127 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


User currently offlineJ_Hallgren From United States of America, joined Jun 2000, 1507 posts, RR: 0
Reply 3, posted (9 years 4 months 3 weeks 3 days 20 hours ago) and read 2123 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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User currently offlineSLC1 From , joined Dec 1969, posts, RR:
Reply 4, posted (9 years 4 months 3 weeks 3 days 20 hours ago) and read 2118 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]

User currently offlineFLYtoEGCC From United Kingdom, joined Feb 2004, 947 posts, RR: 2
Reply 5, posted (9 years 4 months 3 weeks 3 days 20 hours ago) and read 2112 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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User currently offlineRedngold From United States of America, joined Mar 2000, 6907 posts, RR: 44
Reply 6, posted (9 years 4 months 3 weeks 3 days 20 hours ago) and read 2108 times:

Argggh I didn't read the question well enough... Definitely need the sines and tangents here...


Up, up and away!
User currently offlineJ_Hallgren From United States of America, joined Jun 2000, 1507 posts, RR: 0
Reply 7, posted (9 years 4 months 3 weeks 3 days 20 hours ago) and read 2106 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!  Smile

[Edited 2005-05-31 15:58:48]


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