| 3 |
| 5 |
| π |
| 2 |
又由知B＜∠ADC可得B＜
| π |
| 2 |
由sinB=
| 5 |
| 13 |
| 12 |
| 13 |
又由cos∠ADC=
| 3 |
| 5 |
| 4 |
| 5 |
从而sin∠BAD=sin（∠ADC-B）=sin∠ADCcosB-cos∠ADCsinB=
| 4 |
| 5 |
| 12 |
| 13 |
| 3 |
| 5 |
| 5 |
| 13 |
| 33 |
| 65 |
由正弦定理得
| AD |
| sinB |
| BD |
| sin∠BAD |
所以AD=
| BD•sinB |
| sin∠BAD |
33×
| ||
|
| 5 |
| 13 |
| 3 |
| 5 |
| 3 |
| 5 |
| π |
| 2 |
| π |
| 2 |
| 5 |
| 13 |
| 12 |
| 13 |
| 3 |
| 5 |
| 4 |
| 5 |
| 4 |
| 5 |
| 12 |
| 13 |
| 3 |
| 5 |
| 5 |
| 13 |
| 33 |
| 65 |
| AD |
| sinB |
| BD |
| sin∠BAD |
| BD•sinB |
| sin∠BAD |
33×
| ||
|