several ways to solve general triangles

1. “Solving General Triangles”
Prepared by:
Aveen Salam

2. Sine Formula : a/(Sin A) = b/(Sin B) = c/(Sin C)
RULE:

3. A
B
C
The result of
the unknown
angle
A B=70° C=?
A+B+C= 180°
60+70+C=180°
C= 50°
Example:
8
Find the unknown angle.
=60°

4. 8
A
B
C
Example: Find side B
=
b
Sin:70° Sin:60°
8 Sin 70° = b Sin60°
Sin Sin
60° 60°
B=8.68
8

5. Example: Find side C
A
B
C
a
=
c
Sin A Sin C
8 = c
Sin Sin
60° 50°
8 Sin 50° = C Sin60°
Sin Sin 60°
60°
C=7.07
8

6. The law of cosines:
𝑎2 = 𝑏2 + 𝑐2 − 2𝑏𝑐𝐶𝑜𝑠A
𝑏2
= 𝑎2
+ 𝑐2
− 2𝑎𝑐𝐶𝑜𝑠𝐵
𝑐2 = 𝑎2 + 𝑏2 − 2𝑎𝑏𝐶𝑜𝑠𝐶

7. Example 2: solve the triangle
First we should find side a by the law of cosines
𝑎2 = 𝑏2 + 𝑐2 − 2𝑏𝑐𝐶𝑜𝑠A
𝑎2=52 + 72 − (2 × 5 × 7 × cos 49)
𝑎2
=74−(70 × 0.6560)
𝑎2
= 74 − 45.924
𝑎2
= 28.075
𝑎2 = 28.075
𝑎 = 5.30

8. After finding side a, we will find angle B then angle C
sin 𝐵
𝑏
=
sin 𝐴
𝑎
sin 𝐵
5
=
sin 49°
5.3
→
sin 𝐵
5
= 0.142
sin 𝐵 =0.142× 5 → sin 𝐵 =0.7122
𝐵 = sin−1
0.7122
𝐵 = 45.4°
𝐴 + 𝐵 + 𝐶 = 180°
𝐶 = 180° − 49° − 45.4°
𝐶 = 85.6°

9. Example 3: solve the triangle
sin 𝐶
𝑐
=
sin 𝐵
𝑏
sin 𝐶
13
=
sin 31°
8
→
sin 𝐶
13
= 0.0643
sin 𝐶 = 0.0643 × 13 → sin 𝐶 = 0.8369
𝐶 = sin−1 0.8369
𝑪 = 𝟓𝟔. 𝟖°
First we will find angle C then angle A
𝐴 + 𝐵 + 𝐶 = 180°
𝐴 = 180° − 31° − 56.8°
𝑨 = 𝟗𝟐. 𝟐°

10. Now we can use the law of sines again to find a
𝑎
sin 𝐴
=
𝑏
sin 𝐵
𝑎
sin 92.2°
=
8
sin 31°
→
𝑎
sin 92.2
= 15.53
𝒂 = 𝟏𝟓. 𝟓𝟐°

11. The law of cosines angle version
cos 𝐴 =
𝑏2
+ 𝑐2
− 𝑎2
2𝑏𝑐
cos 𝐵 =
𝑐2 + 𝑎2 − 𝑏2
2𝑐𝑎
cos 𝐶 =
𝑎2 + 𝑏2 − 𝑐2
2𝑎𝑏

12. Example 4: solve the triangle
cos 𝐴 =
𝑏2 + 𝑐2 − 𝑎2
2𝑏𝑐
cos 𝐴 =
62 + 72 − 82
2 × 6 × 7
cos 𝐴 =
21
84
→ cos 𝐴 = 0.25
𝐴 = cos−1
0.25
𝑨 = 𝟕𝟓. 𝟓°

13. We will use the law of cosines again but this time for angle B
Then find angle C
cos 𝐵 =
𝑐2 + 𝑎2 − 𝑏2
2𝑐𝑎
cos 𝐵 =
72
+ 82
− 62
2 × 7 × 8
cos 𝐵 =
77
112
→ B = cos−1 0.6875
𝑩 = 𝟒𝟔. 𝟔
𝐴 + 𝐵 + 𝐶 = 180°
𝐶 = 180° − 75.5° − 46.6°
𝑪 = 𝟓𝟕. 𝟗°

14. THANK YOU