(1)在Rt△ABC中,有BC=AB÷tanα=
| h |
| tanα |
同理:在Rt△ABD中,有BD=AB÷tanβ=
| h |
| tanβ |
且CD=BC-BD=m;即
| h |
| tanα |
| h |
| tanβ |
故h=
| m•tanα•tanβ |
| tanβ−tanα |
(2)将α=45°,β=60°,m=50米,代入(1)中关系式可得
h=
| 50米×tan45°•tan60° |
| tan60°−tan45° |
=
50米×1×
| ||
|
=75米+25
| 3 |
≈118.3米.

| 2 |
| 3 |
| h |
| tanα |
| h |
| tanβ |
| h |
| tanα |
| h |
| tanβ |
| m•tanα•tanβ |
| tanβ−tanα |
| 50米×tan45°•tan60° |
| tan60°−tan45° |
50米×1×
| ||
|
| 3 |