town q is 3okm due north of p the bearing of town R from Q is 130degree if R is 10km from Q calculate the bearing of R from P to the nearest degree and how far north of P,R is ,to 2significant number

Question

Bearing  world problems

Answer 1

 

SR=10sin50 (SOH) and QS=10cos50 (CAH). So PS=30-QS=30-10cos50=23.57km. tanQPR=SR/PS=10sin50/(30-10cos50)=0.3250 (TOA), so QPR=18º approx., which is the bearing of R from P. PS=23.57km is the distance of R to the north of P. PR is the actual distance between P and R. SR/PR=sin18 (SOH), so PR=10sin50/sin18=24.79km. (SOHCAHTOA).