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A.MATHS QUESTION~~~~~~~~~~~~
發問:
Q and R are 9 km apart and are both due east of O. The bearing of P and O is SBE, where 0
最佳解答:
Consider DPQR. By cosine law, cosPRQ = (QR2 + PR2 - PQ2)/(2?QR?PR) = (92 + 82 - 62)/(2 x 9 x 8) = 0.7569 PRQ = 40.8o Consider DPOR. By sine law: (sinPOQ)/8 = (sinPRQ)/15 sinPOQ = 8(sin40.8o)/15 sinPOQ = 0.3485 POQ = 20o B = 90o - 20o = 70o =
A.MATHS QUESTION~~~~~~~~~~~~
發問:
Q and R are 9 km apart and are both due east of O. The bearing of P and O is SBE, where 0
最佳解答:
Consider DPQR. By cosine law, cosPRQ = (QR2 + PR2 - PQ2)/(2?QR?PR) = (92 + 82 - 62)/(2 x 9 x 8) = 0.7569 PRQ = 40.8o Consider DPOR. By sine law: (sinPOQ)/8 = (sinPRQ)/15 sinPOQ = 8(sin40.8o)/15 sinPOQ = 0.3485 POQ = 20o B = 90o - 20o = 70o =
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