Question

3 С 130 components parallel to OB and OC. Resolve the position vector A of the car (measured from fixed point) into answer: 2210 ft along o8 2570 ft alona 0C 65° 750-C Please explain in detail --」 3500 ft

Answer 1

We can find magnitude of A vector theta = tan^-1 (20/35) theta = 29.74 degree in triangle OBD = squareroot 2000^2 + 3500^2 = 4031.13 ft Applying sin rule in triangle OBD: sin(35.26)/BD = sin(115)/4031.13 = sin(29.74)/OB BD = 2567.7 ft Ans along OC OB = 4031.13 sin(29.74) OB = 2206.44 ft Ans along OB.