## Sine rule, Cosine rule & Area

Aims: By the end of this chapter, you will be able to
(i) use sine and cosine rules to solve basic trigonometric questions,
(ii) use the area formula with sine, and
(iii) provide proofs to the above rules and formula.

### Sine rule

For any triangle as in diagram 5, we obtain
[a/sin A] = [b/sin B] = [c/sin C] or
(sin A)/a = (sin B)/b = (sin C)/c
This is a handy formula for solving a triangle when given 2 angles and a side or 2 sides and an angle. Note that the latter conditions may give rise to an ambiguous case.
 Example 4 Refer to diagram 5 above. Let angle A and angle B be 38o and 93o respectively. Side b is 51.7 cm. Find side a, side c, and ∠ C. Solutions: Angle C = 180o - 93o - 38o = 49o c/sin C = b/sin B c = (b sin C) /sin B c = (51.7 sin 49o) /sin 93o c ≈ 39.1 cm (3 s.f.) a = (b sin A) / sin B a = (51.7 sin 38o) /sin 93o c ≈ 31.9 cm (3 s.f.) Example 5 Refer to diagram 5 above. Let a=30.9 cm , c= 27.7 cm and ∠ A=75o. Find angle B, angle C and side b. Solutions: sin C/c = sin A/a sin C = (c sin A) / a C = arcsin[ (c sin A)/a ] C = arcsin[ (27.7sin 75o)/30.9 ] C ≈ 60.0o (3 s.f.) ∠ B = 180o- 75o - 60.0o ≈ 45o (3 s.f.) b/sin B = a/sin A b = (a sin B)/sin A b = (30.9 sin 45o)/sin75o b ≈ 22.6 cm (3 s.f.)

The Ambiguous Case.

When we have two sides and an angle then we might not always have unique solution. Let us illustrate this using an example. Let ABC be a triangle. AC and BC are respectively 18 cm and 11 cm. Angle CAB (also written as CÂB or∠ "CAB ) is 28o. Solve the triangle accurate to 1 decimal place.
Let us first draw this triangle:

1. Fix a point say A on a very long horizontal line.
2. At point A form an angle that is 28o.
3. Measure out 18 cm for side AC.
4. Take a compass and from C form an arc that has 11 cm radius.
5. Note that the arc intersects the horizontal line in two places. These intersections are the possible points for B. The intersection furthest to the right form an angle B which is x degree.
sin B/b =sin A/a
sin x = (b sin A)/a
x = arcsin [ (b sin A)/a ]
x = arcsin [ (18 sin 28o)/11 ]
x ≈ 50.2o (1 d.p)
Angle ABC ≈ 50.2o (1 d.p) Thus, angle ACB = 101.8o and AB = 22.9m.
B could also be at the intersection furthest to the left. Angle ABC then is simply 180o - 50.2o or 129.8o. Then angle ACB = 22.2o and AB = 8.9 cm.
Please confirm all the above calculations.

### Cosine rule.

We can think of the cosine rule as an adjustment (the bolded part of the formula) that we need to make to the pythagoras theorem when we don't have a right angle triangle. The cosine rule has three version. We again refer to diagram 5 above.
a2 = b2 + c2 - 2 bc cos A.
b2 = a2 + c2 - 2 ac cos B.
c2 = a2 + b2 - 2 ab cos C.
In general: (side3)2 = (side1)2 + (side2)2 - 2 (side1)(side2) cos (angle described by side1 and side2).
The rule is useful when we have three sides or two sides with one side forming the known angle. The above expression can be rewritten into A = arccos [ (b2 + c2 - a2) / ( 2bc ) ] (see the box at the bottom of the page for a TI programme.)
 Example 6. Refer to diagram 5. Let b=11.2 cm, c= 16.2 and angle A= 63o. Find side a and use your answer for side a to find angle B. Solutions: a2 = b2 + c2 - 2 bc cos A. a2 = 11.22 + 16.22 - 2(11.20)(16.2) cos 63o. a ≈14.9 cm (3 s.f.) sin B/b = sin A/a sin B = (b sin A)/a B = arcsin [ 11.2sin63o/(14.9) ] B ≈ 41.9o Example 7. Refer to diagram 5. Let a= 13.6, b= 13.6 and c=16.3. Find angle C. Solution: C = arccos [ (a2 + b2 - c2) / ( 2ab ) ]. C = arccos [ (20.72 + 16.32 - 13.62) / ( 2(16.7)(16.3) ) ]. C ≈ 41.0o

### Area of a triangle.

The area of any triangle such as in diagram 5 can be found by using
Area = (1/2)bc sin A or
Area = (1/2)ac sin B or
Area = (1/2)ab sin C
Basically the area of a triangle is just (1/2)(side1)(side2) sin(angle formed by side1 and side2) .
 Example 8. Find the area of the triangle in example 6 and example 7 above. Solutions: Area (example 6) = (1/2)bc sin A We have picked an abitrary angle A for this soultion but we will arrive at the same answer if we would have chosen other sides and angle. Area = (1/2)(11.2)(16.2) sin 63o Area ≈ 80.8 cm2 (3 s.f.) Area (example 7) = (1/2)ab sin C Area = (1/2)(20.7)(13.6) sin 41.0o Area ≈ 92.3 cm2 (3 s.f.)

### Some algebraic proofs.

Show that the area of a triangle ABC as in diagram 5 is Area = (1/2)bc sin A.
A triangle ABC could be segmented into two right angle triangles as in diagram 6.
Area = (1/2)bh
h = c sin A. [use the identity of sin A]
Thus, Area = (1/2)bc sin A. Similar argument can be used to show
Area = (1/2)ac sin B and
Area = (1/2)ab sin C.
Prove the last two area expressions.

### Deriving the sine rule.

Since (1/2)bc sin A = (1/2)ac sin B and (1/2)bc sin A = (1/2)ab sin C
then
bc sin A = ac sin B = ab sin C
Multiply the whole thing by 2/(abc)
sin A/a = sin B/b = sin C/c
Let us rewrite the above as sin A/a = sin B/b and sin A/a = sin C/c then b/sin B = a/ sin A and c/sin C = a/sin A.
Thus, we obtain
a/sin A = b/sin B = c/sin C.

### Deriving the cosine rule.

Refer to diagram 6 above. Let h be a line that drops perpendicularly from corner C to side AB. Let M be the point of intersection between h and AB.
Let the length AM be x and x = b cos A.
From the two right angle triangles, we obtain the following:
a2 = h2 + (c - x)2 and
b2 = h2 + (x)2
Thus,
a2 - c - x)2 = b2 - (x)2
a2 = b2 - (x)2 + (c - x)2
a2 = b2 - (x)2 + c2 + (x)2 - 2b(x)
a2 = b2 + c2 - 2b(x)
a2 = b2 +c2 - 2bc cos A