Mathematics

ai)
OB² = 300² + 100² - 2*300*100 cos(90 - 30)°
OB² = 100000 - 30000
OB = 100√7 maii)
tan∠TAO = TO / 300
tan∠TAO = OB tan35° / 300
tan∠TAO = 100√7 tan35° / 300
tan∠TAO = 0.61753
∠TAO = 31.69637 = 32°(nearest degree)
b)
BC = 300 - 100sin30° = 250 m
ci)
sin∠OBA / 300 = sin∠OAB / OB
sin∠OBA / 300 = sin60° / (100√7)
sin∠OBA = (3/2)√(3/7) (BD/2) / OB = sin∠OBA
(BD/2) / (100√7) = (3/2)√(3/7)
BD = 300√3 AD = BD - BA = 300√3 - 100 = 420 m (3 sig. fig.)
cii)
By ci) , sin∠OBA = (3/2)√(3/7)
∠OBA = 79.1°∠OBC = 180° - 79.1° - 60° = 40.9° So ∠OBA > ∠OBC , then BD < BE , Peter route is shorter.