Question

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16 Gi Given the following figure resulting from a radial traverse with "O" as the observation point in the middle, what is: a. length AF, in ft (10 points)? b. the angle BAF (angles BAO and FAO collectively), in DD MM SS (10 points)? 200 ft 200ft 1 7080 V 150 ft 60° 150 ft V 100 ft Eb - 2012(a) coscos

Answer 1

Solution. no • 200ft 200 dt 150ft 60° 60 609 150H taost tov fig(e) Ер To solve this problem we have to use the sine rule and cosine rule sine Rule: San ABC is a triangle here (fig (2)] Accoording to sine Rule; B2 AC AB Sim A Sim B sine ar sana sing sine where LA, LB, LC are the angles and a, b, c are the side lengths opposite to those angles respectively. Bc

Cosine Rule: Respect to fig (2); Say we know the value of all three sides. Then by using this side lengths we can find the angles. The method known as Cosine Rule. COSA- 6² c²a2 250 Cos B = a + c² - 6² zac Со с а”ь"- 2 ab Now het as back to our problem; 6 (a) Here we have to find the length of AF in ft, So in A AOF; AO=200ft - For 150 ft LAOF = 60° using cosine gule here and so we can write; COSLAOF = 40 + F0²_ AF2. алютно oh, Cosco = 200² +150% - A F² 2x200x150 09, AK 200% +1502 - 2x 20041 cox cos 60) or AF-532500 - or AF- 180.2776 ft

(6 Now we have to find 2BAF by finding LBAO and LFAO at first. In A Or if we use sine rull; then we can write - AF - Ao For sin LAOF Sin LAFO - SinLIAO 180. 2776 200 sim 69. Sin LAFO sinp - 150 - 180.2776 1.50 singoo - sing oh, Sing= 150x singoo (80.2776 o, ß-sin/ 150xsino a 4686'08" 180.27th d And from A ROB; we can see here; 2+55° + 70°= 180° 09, d=188-559-7° or, &- 55°00'00" Therefore ZBAF = LBAO + LFAO LBAF = 2 + 3 LBAF - 5500000" + 46 06'08" ZBAF = 101 0608" (Answer)