Question

7 This diagram shows a straight track through points A, S and B, where A is 10 km northwest of B and S is exactly halfway between A and B. A surveyor is required to reroute the track through P from A to B to avoid a major subsidence at S. The surveyor determines that A is on a bearing of 330 from P and that B is on a bearing of 070° from P. Assume the region under consideration is flat. Find: a the magnitudes of angles APB. PAB and PBA b the distance from P to B and from P to S S the bearing of S from P d the distance from A to B through P. if the surveyor chooses to reroute the track along a circular arc. .P B

Answer 1

Soln C®) magnitude - of LAPB = 3i+70=100 A Given AB = 10km. AS=5B=5 km S 45' Bot 1070 zo A is North west of B 19045 But Angle with horizontal and 330 vertical =45 LABP = 65 LPAB = 180-100-65-15 (6) distance from pto B sinloo APAB =) sinis" (sine Rule) AB PB PB = 10 sinis" 2.628 ~ 2-63 tm sinloo APSB = cosine Rule cos 65 PB²+ Bs²_ps? 2 PB.BS ps²= PB²t6s²2 ws$5) B.BS - (2-63345²-226565 (2-63). Ps²= 20-802039 5 () ps = 4.56. bearing of s from P. sin LSPA ASAP AS SP <SPA sin" (AS Y AS sinis) = sint 15 sinis) 4.56 - 16-48679 s from P = 330+16.49 346-49² SP

(d) LAPB = 100' A APB Circulan Arc. Radius of circular are. >lokm. s loo 10 Rsingolo A Al 100 P À B diameter) o centre Angle sub tend by AB at centre of R Radius circular Arc 160 Radius of circulan Arc = R sin go 5.077 km = Arc length of circulen Arc = 160 T 180 s singo = Hot g sin 180) = 14.178 km.

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