An aircraft flies 400 km from a point O on a bearing of 025° and then 700 km on a bearing of 080° to arrive at B. a) How far north of O is B? b) How far east of O is B? c) Find the distance and bearing of B from O. The best answer will be marked as brainliest!!!
Answers
Step-by-step explanation:
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Answer:
(a). 484.07 km
(b). 858.40 km
(c). Distance =985 km; Bearing of B from O = 60.6 degrees
Step-by-step explanation:
(a). how far north of O is B:
cos ∅= adj 1/hyp 1
cos 80° = adj / 700
adj 1 = 700 (cos 80°)
adj 1 = 121.55 km
cos ∅ = adj 2/ hyp 2
cos 25° = adj 2 / 400
adj 2 = 400 (cos 25°)
adj 2 = 362.523
adj = adj 1 + adj 2
adj = 121.55 + 362.523
= 484.07
distance covered over north = 484 km (approx)
(b). how far east of O is B:
sin∅ = opp 1 / hyp 1
sin 80° = opp 1 / 700
opp 1 = 700 (sin 80°)
opp 1 = 689.365 km
sin∅ = opp 2 / hyp 2
sin 25° = opp 2 / 400
opp 2 = 400 (sin 25°)
opp 2 = 169.04 km
opp = opp 1 + opp 2
= 689.365 + 169.04
opp = 858.40 km
distance covered over east = 858 km (approx)
(c). the distance and bearing of B from O
distance → OB =d
d² = (opp)² + (adj)²
= (858)² + (484)²
d = 985.09 km
The distance = 985.09 km
bearing of B from O
tan ∅ = opp / adj
tan ∅ = 858 / 484
∅ = tan⁻¹ (1.722)
∅ = 60.57°
Bearing of B from O is 60.6° (approx)