由正弦定理及sinB＝
| 5 |
| 13 |
| 25 |
| 169 |
| 1 |
| tanA |
| 1 |
| tanC |
| cosA |
| sinA |
| cosC |
| sinC |
| sin(A+C) |
| sinAsinC |
| sinB |
| sinAsinC |
| 5 |
| 13 |
| 169 |
| 25 |
| 13 |
| 5 |
（2）由accosB=12知cosB＞0．
由sinB＝
| 5 |
| 13 |
| 12 |
| 13 |
从而，b2＝ac＝
| 12 |
| cosB |
由余弦定理，得b2=（a+c）2-2ac-2accosB．
代入数值，得13＝(a+c)2−2×13×(1+
| 12 |
| 13 |
解得：a+c＝3
| 7 |
| 5 |
| 13 |
| 1 |
| tanA |
| 1 |
| tanC |
| 5 |
| 13 |
| 25 |
| 169 |
| 1 |
| tanA |
| 1 |
| tanC |
| cosA |
| sinA |
| cosC |
| sinC |
| sin(A+C) |
| sinAsinC |
| sinB |
| sinAsinC |
| 5 |
| 13 |
| 169 |
| 25 |
| 13 |
| 5 |
| 5 |
| 13 |
| 12 |
| 13 |
| 12 |
| cosB |
| 12 |
| 13 |
| 7 |