The law of Sines or Sine Rule

The sides of a triangle are proportional to the sines of the angles opposite to them i.e.
   a/ sin A = b/sin B = c/sin C

Napier Analogies (Laws of tangents)

In any ABC,

1. tan (B -C)/2 = [(b -c)/(b+c)]cot A/2
2. tan (C -A)/2 = [(c - a)/(c+a)]cot B/2
3. tan (A -B)/2 = [(a -b)/(a+b)]cot C/2

Law of cosines or cosine formula

In any ABC,

1. cos A = (b² +c² -a²)/2bc
2. cos B = (a² +c² -b²)/2ca
3. cos C = (a² +b² -c²)/2ab

Projection formulae

In any triangle ABC,

1. a = b cos C +c cos B
2. b = a cos C +c cos A
3. c = a cos B +b cos A

Illustrative Examples

Example

Using sine formula, prove the cosine formula.

Solution

Let ABC be any triangle. Using sine rule,
    a/sin A = b/sin B = c/sin C = k (say)
=> a = k sin A, b = k sin B, c = k sin C
Now a² +b² - c² = k² sin² A + k² sin² B -k² sin² C
= k² sin² A + k² (sin² B -sin² C)
= k² sin² A + k² sin (B +C) sin (B -C)
= k² sin² A + k² sin A sin (B -C)
         [ A +B +C =   =>  B +C = -A   =>  sin (B +C) = sin ( -A) = sin A]
= k² sin A [sin A +sin (B -C)]
= k² sin A [sin (B +C) +sin (B -C)] = k² sin A (2 sin B cos C)
= 2 (k sin A)(k sin B) cos C = 2 a b cos C
=>    cos C = (a² +b² -c².)/2ab
Similarly the other two cosine formulae can be proved.

Example

In a ABC right angled at B, prove that

1.  
2. tan (C -A)/2 = (c-a)/(c+a)

Solution

1. By Napier analogy, tan (B -C)/2 = [(b -c)/(b+c)] cot A/2           ....(1)
   If B = 90°, we get A + C = 90°
   =>   B - C = 90° -(90° - A) = A
   => tan (B-C)/2 = tan A/2
   Hence from (1), we get tan A/2 = [(b-c)/(b+c)].[1/tan A/2]
   tan²A/2= (b-c)/(b+c)
   So tan A/2 = (Taking +ve value as A/2 < 90°)
2. By Napier analogy, tan (C -A)/2 = [(c -a)/(c+a)] cot B/2
          = [(c -a)/(c+a)]cot 90°/2 = [(c -a)/(c+a)]

Example

If a cos A = b cos B, then prove that either the triangle is isosceles or right-angled.

Solution

Let a/sin A = b/sin B = c/sin C = k (say)
=>   a = k sin A, b = k sin B, c = k sin C
Now a cos A = b cos B   =>   k sin A cos A = k sin B cos B
=> 2 sin A cos A = 2 sin B cos B   =>  sin 2 A = sin 2 B
=> sin 2 A -sin 2 B = 0   => 2 cos (A +B) sin (A -B) = 0
=> cos (A +B) = 0 or sin (A -B) = 0
=> either A +B = 90° or A -B = 0° (0° < A +B < 180°)
=> C = 180° -(A +B) = 180° -90° = 90° or A = B
=> triangle is rt d or triangle is isosceles

Exercise

In any triangle ABC, prove the following (1 -14):

1. cos (B -C)/2 = 2 sin A/2, where b +c = 2 a
2. (c -b cos A)/(b -c cos A) = cos B/cos C
3. (a +b) cos C +(b +c) cos A +(c +a) cos B = a +b +c
4. (c -b)² cos² A/2 +(c +b)² sin² A/2 = a²
5. a²sin (B -C)/[sin B +sin C ] +b²sin (C -A)/[sin C +sin A] + c² sin (A - B)/[sin A +sin B] = 0
6. (a² -b²)/(cos A +cos B) + b² -c²)/(cos B +cos C) + (c² -a²)/(cos C +cos A) = 0
7. b² sin 2 C +c² sin 2 B = 2 b c sin A
8. [(b² -c²)/a] cos A +[(c² -b²)/b] cos B + [(a² -b²)/c] cos C = 0
9. (cos A)/a + (cos B)/b + (cos C)/c = (a² +b² +c²)/(2abc)
10. b² cos 2 A -a² cos 2 B = b² -a²
11. a cos (A +B +C) -b cos (B +A) -c cos (A +C) = 0
12. (i) sin B +sin C > sin A.
    (ii) (sin A +sin B)(sin B +sin C)(sin C +sin A) > sin A sin B sin C
    [Hint. Use a +b > c etc.]
13. (c² -a² + b²) tan A = (a² - b² +c²) tan B = (b² - c² +a²) tan C.
14. sin³ A cos (B -C) +sin³ B cos (C -A) +sin³ C cos (A -B) = 3 sin A sin B sin C.
15. If in a triangle ABC, sin 2 A +sin 2 B = sin 2 C, prove that either A = 90° or B = 90°.
16. If in a triangle ABC, sin² A +sin² B = sin² C, prove that ABC is right angled.
17. The sides of a triangle are of lengths 5 x +4 y, 4 x +5 y and 7 x +7 y units, where x and y are positive. Show that the triangle is obtuse.
18. If the angles of a triangle are in the ratio 1: 2 :3, prove that the corresponding sides of the triangle are in ratio 1 : 3 2.
19. In a ABC, a = 1, b = and C = /6. Find the other two angles and the third side.
20. If the angles A, B, C of a triangle ABC are in the ratio 3:4:4 prove that a +c2 = 2 b.
21. In a ABC, the sides are 3, 5, 7 units. Using cosine formula, find the angles in degree measure (upto one decimal).
22. If a = 29· 45 cm, b = 30·12 cm and B = 66°, find the remaining side and angles of the triangle.
23. If a = 2· 0 ft, b = 3·0 ft, A = 23·1° and B is obtuse, show that c = 1·1 ft.
24. Show that there is no triangle satisfying the conditions a = 2·0 ft,b = 6·0 ft and A = 23·1°.
    [Hint. Try computing sin B using the law of sines.]
25. Let b = 1, a = 2 and B = 30°. Show that A = 45° or 135°.
    (i) Taking A = 45°, determine the remaining parts of ABC.
    (ii) Taking A = 135°, determine the remaining parts of ABC.

Answers