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4. The sine and cosine rules

In trigonometry the convention is to use lower-case letters for the sides of a figure, and upper-case letters for the angles. One should also be aware that one should denote the side of a figure according to the letter denoting the opposite angle (if the opposite angle is "A", then the side will be "a"). One should be aware of this as it is integral to the understanding of the sine and cosine rules.

The area of an acute-angled triangle

As one should be already aware, in a triangle:

 Area \ = \frac{b \times h}{2}

Where "b" is the base, and "h" is the perpendicular height.

This formula is easy to use in a triangle where one is aware of the perpendicular height (right-angled triangles are often like this [due to the right-angled within the triangle]), however in other triangles it is a bit more difficult.

Image:Area_Of_A_Triangle.PNG

Figure 001

The above image shows how one can construct a perpendicular line, which cuts the acute-angled triangle into two right-angled triangles, which one can work with. The idea is to obtain the value of "h" (the perpendicular height), from which one is able to use the original formula such that one can obtain the area of the triangle.

Consider that:

 \sin{B} = \frac{Opposite}{Hypotenuse} = \frac{h}{c}

Similarly:

 \sin{C} = \frac{Opposite}{Hypotenuse} = \frac{h}{b}

Taking the latter:

 h = b \times \sin{C}

Hence:

 Area \ = \frac{a \times h}{2} = \frac{1}{2} \times a \times b \times \sin{C}

Area of an obtuse-angled triangle

Now one must consider the case of the triangle being an obtuse-angled triangle.

Image:Area_Of_A_Triangle_(Obtuse).PNG

Consider now that the angle which is made between the line AC, and the extension of BC (in the context of the acute angle) is  180^{ \circ } - C .

Now:

 \sin{180^{ \circ } - C} = \sin{C} = \frac{Opposite}{Hypotenuse} = \frac{h}{b}

Hence:

 h = b \times \sin{C}

Hence, the area can now be dealt with:

 Area \ = \frac{1}{2} \times a \times b \times \sin{C}

Hence, one can conclude that for all triangles:

 Area \ = \frac{1}{2}ab \sin{C} .

One must note that the fact that there are two sides multiplied by the sine of the angle between them, and then it is halved. It is important that one picks appropriate sides, and not just any set of two sides and an angle.

The sine rule for a triangle

One can rewrite the formula for the area of any triangle thus:

 \frac{1}{2}ab \sin{C} = \frac{1}{2}bc \sin{A} = \frac{1}{2}ac \sin{B}

Hence:

 ab \sin{C} = bc \sin{A} = ac \sin{B}

Now one can divide by "abc":

 \frac{ \sin{C}}{c} = \frac{ \sin{A}}{a} = \frac{ \sin{B}}{b} .

This is an important formula, and can be used to calculate the angles, and sides of a triangle.

Example

1. Study the below figure:

Image:Triangle.PNG

This triangle has a side length of  \sqrt{3} units.

Calculate the area.

 Area \ = \frac{1}{2}ab \sin{C} = \frac{1}{2} \times \left( \sqrt{3} \right) ^{2} \times \sin{60^{ \circ }}

Therefore:

 Area \ = \frac{3}{2} \times \frac{ \sqrt{3}}{2} = \frac{3}{4} \sqrt{3}

One must also note that the formula can be rearranged:

 \frac{a}{ \sin{A}} = \frac{b}{ \sin{B}} = \frac{c}{ \sin{C}}

The longest and shortest sides of a triangle

One should be aware that the longest side of a triangle is always opposite the largest angle, and the smallest side is always opposite the smallest angle of the triangle.

Consider a triangle ABC, where:  A > B .

If  A, \ B < 90^{ \circ } :

Consider the graph of:  y = \sin{x} , for  0^{ \circ } < x < 90^{ \circ } .

Now one can see that if  A > B ,  \sin{A} > \sin{B} (as the gradient is always positive in this region).

If angle A is obtuse:

Due to the angles in a triangle summing to  180^{ \circ } ,  A + B < 180^{ \circ} . Hence:

 180^{ \circ } - A > B

Hence:

 \sin{A} > \sin{B}

(As  \sin{ \left( 180^{ \circ } - A \right)} = \sin{A} ).

If  \sin{A} > \sin{B} , let:  \sin{A} = \sin{B} + \epsilon .

Hence:

 \displaystyle \frac{ \sin{A}}{a} = \frac{ \sin{B}}{b} = \frac{ \sin{B} + \epsilon}{a}

 \displaystyle \frac{ \sin{B}}{b} = \frac{ \sin{B} + \epsilon}{a}

 \displaystyle b = \frac{ \sin{B}}{ \frac{ \sin{B} + \epsilon}{a}}

 \displaystyle b = \frac{ a \sin{B}}{ \sin{B} + \epsilon}

 \displaystyle b = a \left[ \frac{ \sin{B}}{ \sin{B} + \epsilon} \right]

As  \epsilon > 0 :

 \frac{ \sin{B}}{ \sin{B} + \epsilon} < 1

Hence:

 b < a .

Now suppose that one is given that  a > b . (Note that one only compares two variables, as this can cover all cases).

 Let \ b + \epsilon = a

 \epsilon > 0

Therefore, by the sine rule:

 \displaystyle \frac{ \sin{A}}{a} = \frac{ \sin{A}}{b + \epsilon} = \frac{ \sin{B}}{b}

Hence:

 \displaystyle  \frac{ \sin{A}}{b + \epsilon} = \frac{ \sin{B}}{b}

 \displaystyle \sin{A} = \frac{ \left( b + \epsilon \right) \sin{B}}{b}

 \displaystyle \sin{A} = \sin{B} \left[ \frac{b + \epsilon}{b} \right]

As  b + \epsilon > b ,  \frac{b + \epsilon}{b} > 1 .

Hence:

 \sin{A} > \sin{B} .

If both of the angles are acute, one can tell quite easily from the graph of the sine function that the only way for the sines of the two angles to have the above relationship is for  A > B .

Now, if A is obtuse, and B is acute:

 A + B < 180^{ \circ }

Hence:

 B < 180^{ \circ } - A

 180^{ \circ } - A < 90^{ \circ } ,

and therefore, as A is obtuse ( > 90^{ \circ } ),  A > B .

If B was obtuse, and A was acute:

If  \sin{A} > \sin{B} ,  \sin{A} > ( \sin{180^{ \circ }}- B) , and  A > (180^{ \circ } - B) .

This is impossible as:

 A + B < 180^{ \circ } ,

and hence:

 A < (180^{ \circ } - B) .

One can see from these results that if  a > b ,  A > B .

The cosine rule for a triangle

If one is aware of the lengths of two sides, and the angle between them, the sine rule cannot give an answer. The sine rule will leave an unknown in each of the fractions, making the problem unsolvable by this method. This is where one needs the cosine rule.

Allow Figure 001, and Figure 002 to be placed upon the Cartesian plane, and the vertex denoted by C is at the origin (0, 0).

One can now rotate the triangle about the origin such that the vertex denoted by A is on the positive x-axis (coordinates (b, 0)).

Now one can see that the B coordinate will be  \left( a \cos{C}, \ a \sin{C} \right) .

One can now use the idea of finding the length of a line to find the length of "c":

 c^{2} = \left( a \sin{C} - 0 \right) ^{2} + \left( a \cos{C} - b \right) ^{2} = a^{2} \sin^{2}{C} + a^{2} \cos^{2}{C} - 2ab \cos{C} + b^{2}

Now one can use the Pythagorean identity to show:

 a^{2} \sin^{2}{C} + a^{2} \cos^{2}{C} = a^{2}

Hence:

 c^{2} = a^{2} + b^{2} - 2ab \cos{C} .

Example

1. Let a = 2, b = 2,  C = 60^{ \circ } . Calculate c.

This is a simple matter of putting the numbers into the formula:

 c^{2} = a^{2} + b^{2} - 2ab \cos{C} = 2^{2} + 2^{2} - (2 \times 2 \times 2 \times \cos{60^{ \circ }})

Hence:

 c^{2} = 8 - (8 \times \frac{1}{2}) = 8 - 4 = 4

 c = \pm 2

Hence:

 c = 2

("c" must be positive due to the context of the question).

Solving triangles

One might be asked to solve a triangle, this is a simple process (generally) of finding the figures which are equivalent to the lengths of the sides, and the angles of the triangle.

In order to complete (correctly) one of these questions one should have a good understanding of the cosine and sine rules, and be confident that one is able to use letters other than A, B, and C with the rules.

One tip that one might find useful is to ensure that one is very accurate (many decimal places in the case of a not obviously recurring decimal) before the final result. Premature rounding can cause error later in the calculation, and result in loss of marks.

One does not have to find all three angles using the sine rule, if one has two angles, one can quite easily cite that the result is a number due to the fact that the sum of the interior angles in a triangle is  180^{ \circ }


Also See

Read these other OCR Core 2 notes:

  1. Trigonometry
  2. Sequences
  3. The binomial theorem
  4. The sine and cosine rules
  5. Integration
  6. Geometric sequences
  7. Exponentials and logarithms
  8. Factors and remainders
  9. Radians
  10. The trapezium rule
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