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圖片參考：http://imgcld.yimg.com/8/n/HA00437808/o/701109150077613873457790.jpg Thanks a lot~
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11. In ΔTYZ : tan27° = (3m)/YZ YZ = 3/tan27° In ΔXYZ : sinDYXZ / YZ = sin DXZY / XY (sine law) sin12° / (3/tan27°) =sinDXZY / 24 sinDXZY = 24 sin12° tan27° /3 DXZY = 57.9° 57.9° +12° =69.9° Bearing of T (Y) from Z = S 69.9° W 14. (a) DBXY = 90° - 60° =30° DBYX = 90° - 30° =60° DXBY + DBXY + DBYX = 180° (D sum of Δ) DXBY + 30° + 60° =180° DXBY = 90° (b) In rt.D ΔBXY : sinDBXY = BY / XY sin30° =BY / (24 km) BY = 24 sin30° km BY = 12 km In rt.D ΔABY : tanDAYB = AB / BY tan10° = AB/ (12 km) AB = 12 tan10° km AB = 2.16 km The height of the aeroplane A above B = 2.16km (c) In rt.D ΔBXY : cosDBXY = BX / XY cos30° =BX / (24 km) BX = 24 cos 30° km In rt.D ΔABX : tanDAXB = AB / BX tanDAXB = (12 tan10°) / (24 cos30°) DAXB = 5.81° The angle of elevation of the aeroplane A from X = 5.81°
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f.5 trigonometry (urgent)發問:
圖片參考：http://imgcld.yimg.com/8/n/HA00437808/o/701109150077613873457790.jpg Thanks a lot~
最佳解答:
11. In ΔTYZ : tan27° = (3m)/YZ YZ = 3/tan27° In ΔXYZ : sinDYXZ / YZ = sin DXZY / XY (sine law) sin12° / (3/tan27°) =sinDXZY / 24 sinDXZY = 24 sin12° tan27° /3 DXZY = 57.9° 57.9° +12° =69.9° Bearing of T (Y) from Z = S 69.9° W 14. (a) DBXY = 90° - 60° =30° DBYX = 90° - 30° =60° DXBY + DBXY + DBYX = 180° (D sum of Δ) DXBY + 30° + 60° =180° DXBY = 90° (b) In rt.D ΔBXY : sinDBXY = BY / XY sin30° =BY / (24 km) BY = 24 sin30° km BY = 12 km In rt.D ΔABY : tanDAYB = AB / BY tan10° = AB/ (12 km) AB = 12 tan10° km AB = 2.16 km The height of the aeroplane A above B = 2.16km (c) In rt.D ΔBXY : cosDBXY = BX / XY cos30° =BX / (24 km) BX = 24 cos 30° km In rt.D ΔABX : tanDAXB = AB / BX tanDAXB = (12 tan10°) / (24 cos30°) DAXB = 5.81° The angle of elevation of the aeroplane A from X = 5.81°
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