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Sines and Cosines. Question. Please Help? |
Still helping the children. Last few questions THEY say... Thankyou so much for helping me out with these questions! Please put down any formula's or any work so that I may try to understand! Thankyou! The points will go to the best explained answer! Good day! A hiking trail is marked out. The map,maker wants to supply bearings for the leaflet that describe the trail. Calculate the measures of the angles between the paths. Round your answers to the nearest tenth for each trail. Image: http://i184.photobucket.com/albums/x13/m... Hi, Use the Law of Cosines: b虏 = a虏 + c虏 -2ac cos B 17.4虏 = 9.8虏 + 11.2虏 -2(9.8)(11.2) cos B 302.76 = 96.04 + 125.44 - 219.52 cos B 81.28 = -219.52 cos B 81.28/(- 219.52) = cos B -.3703 = cos B B = cos^(-1) (.3703) = 111.7掳 a虏 = b虏 + c虏 -2bc cos A 9.8虏 = 17.4虏 + 11.2虏 -2(17.4)(11.2) cos A 96.04 = 302.76 + 125.44 - 389.76 cos A -332.16 = -389.76 cos A -332.16/(- 389.76) = cos A .8522 = cos A A = cos^(-1) (.8522) = 31.5掳 C = 180 - A - B = 180 - 31.5 - 111.7 = 36.8掳 I hope that helps!! :-) cos B= (a虏 + c虏 - b虏)/ 2ac ...Use when given 3 sides cos B= (9.8虏 + 11.2虏 - 17.4虏)/ 2(9.8)(11.2) B= cos^-1[ (9.8虏 + 11.2虏 - 17.4虏)/ 2(9.8)(11.2) ] B = cos^(-1) (.3703) B = 111.7掳 similarily for other angles cos A= (b虏 + c虏 - a虏)/ 2bc use COS rule a(a)=b(b)+c(c)- 2bc cosA b(b)=a(a)+c(c)- 2ac cosB c(c)=a(a)+b(b)- 2bc cosC a, b,c stand for length n A is angle NOTE:a is opposite to A(in simple words, a is facing A), n so on for b&B and c&C. im not sure which angle u want, so i wil calculate or angle A 9.8(9.8)=17.4(17.4)+11.2(11.2) - 2 (17.4)(11.2) cos A 96.04=302.76+125.44 - 389.76 cos A 96.04-302.76-125.44=-389.76 cos A -332.16=-389.76 cosA 389.76 cos A=332.16 cos A=332.16/389.76 A=31.5464 degree (to change get rid or cuz, use shift pewer negative 1 on 322.16/389.76) angle B 17.4(17.4)=9.8(9.8)+11.2(11.2) - 2 (9.8)(11.2) cos B 302.76=96.04+125.44 - 219.52cos B 302.76-96.04-125.44=-219.52 cos B 81.28=-219.52 cos B cos B=81.28/-219.52 B=111.7318degree angle C 11.2(11.2)=9.8(9.8)+17.4(17.4) - 2 (9.8)(17.4) cos C 125.44=96.04+302.76 - 341.04cos C 125.44-96.04-302.76=-341.04cos C -273.36=-341.04 cos C cos C=-273.36/-341.04 C=36.7218degree check A+B+C=18O degree 34.2530 +111.7318+use COS rule a(a)=b(b)+c(c)- 2bc cosA b(b)=a(a)+c(c)- 2ac cosB c(c)=a(a)+b(b)- 2bc cosC a, b,c stand for length n A is angle NOTE:a is opposite to A(in simple words, a is facing A), n so on for b&B and c&C. im not sure which angle u want, so i wil calculate or angle A 9.8(9.8)=17.4(17.4)+11.2(11.2) - 2 (17.4)(11.2) cos A 96.04=302.76+125.44 - 389.76 cos A 96.04-302.76-125.44=-389.76 cos A -322.16=-389.76 cosA 389.76 cos A=322.16 cos A=322.16/389.76 A=34.2530 degree (to change get rid or cuz, use shift pewer negative 1 on 322.16/389.76) angle B 17.4(17.4)=9.8(9.8)+11.2(11.2) - 2 (9.8)(11.2) cos B 302.76=96.04+125.44 - 219.52cos B 302.76-96.04-125.44=-219.52 cos B 81.28=-219.52 cos B cos B=81.28/-219.52 B=111.7318degree angle C 11.2(11.2)=9.8(9.8)+17.4(17.4) - 2 (9.8)(17.4) cos C 125.44=96.04+302.76 - 341.04cos C 125.44-96.04-302.76=-341.04cos C -273.36=-341.04 cos C cos C=-273.36/-341.04 C=36.7218degree check A+B+C=18O degree 31.5464 +111.7318+36.7218=180 |
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