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.
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 38^{o} and 93^{o} respectively. Side b is 51.7 cm. Find side a, side c, and ∠ C. Solutions: Angle C = 180^{o}  93^{o}  38^{o} = 49^{o} c = (b sin C) /sin B c = (51.7 sin 49^{o}) /sin 93^{o} c ≈ 39.1 cm (3 s.f.) 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=75^{o}. 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 75^{o})/30.9 ] C ≈ 60.0^{o} (3 s.f.) ≈ 45^{o} (3 s.f.) b = (a sin B)/sin A b = (30.9 sin 45^{o})/sin75^{o} b ≈ 22.6 cm (3 s.f.) 
The Ambiguous Case.
INVESTIGATION: Given an angle and two lengths, can we always construct a triangle? Task 1: Let
angle CAB be 60^{o}, AB = 10 cm and CB = 5√ 3. In
this case, you will be able to construct a triangle very similar to
that in diagram 5. Task 2: Let angle CAB be 60^{o}, AB = 10 cm
and CB = 2. Is it possible to construct this triangle? Note: position
C is unknown.
Task 3: Let angle CAB be 60^{o}, AB = 10 cm
and CB = 5. Is it possible to construct this triangle?
Task 4: Let angle CAB be 60^{o}, AB = 10 cm
and CB = 13. Is it possible to construct this triangle?
Task 5: Let angle CAB be 60^{o}, AB = 10 cm
and CB = 9. Is it possible to construct this triangle?
Task6: Let angle CAB be 60^{o}, AB = 10 cm and
CB = k.

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 28^{o}. Solve the triangle accurate to 1 decimal place.
Let us first draw this triangle:
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.
a^{2} = b^{2} + c^{2}  2 bc cos A.
b^{2} = a^{2} + c^{2}  2 ac cos B.
c^{2} = a^{2} + b^{2}  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 [ (b^{2} +
c^{2}  a^{2}) / ( 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= 63^{o}. Find side a and use your answer for side a to find angle B. Solutions: a^{2} = b^{2} + c^{2}  2 bc cos A. a^{2} = 11.2^{2} + 16.2^{2}  2(11.20)(16.2) cos 63^{o}. a ≈14.9 cm (3 s.f.) sin B = (b sin A)/a B = arcsin [ 11.2sin63^{o}/(14.9) ] B ≈ 41.9^{o} 
Example 7. Refer to diagram 5. Let a= 13.6, b= 13.6 and c=16.3. Find angle C. Solution: C = arccos [ (a^{2} + b^{2}  c^{2}) / ( 2ab ) ]. C = arccos [ (20.7^{2} + 16.3^{2}  13.6^{2}) / ( 2(16.7)(16.3) ) ]. C ≈ 41.0^{o} 
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 63^{o} Area ≈ 80.8 cm^{2} (3 s.f.) Area = (1/2)(20.7)(13.6) sin 41.0^{o} Area ≈ 92.3 cm^{2} (3 s.f.) 
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.