Help.........................

Xét `\triangleADC`, có:

`tan\hat{ACD} = (AD)/(DC) `

`=> AD = DC . tan\hat{ACD} = 80. tan60^0 = 80\sqrt{3} (m)`

Xét `\triangleBDC`, có:

`tan\hat{BCD} = (BD)/(DC) `

`=> BD = DC. tan{BCD} = 80. tan45^0 = 80 (m)`

`=> AB = AD - BD = 80\sqrt{3} - 80 ~~58,56 (m)`

Vậy khoảng cách `AB` xấp xỉ `58,56` mét.