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The cosine rule
Determining the missing data of a triangle is not a complicated task in certain cases. For example if the triangle is a right-angled triangle, then the Pythagorean theorem makes a connection between its sides. According to this in the case of a right-angled triangle the sum of squares of the legs is equal to the square of the hypotenuse:
\latex{a^2+b^2=c^2}.
It is obvious that in the case when there is an acute angle opposite the side \latex{ c }, then the equality implies the following inequality:
\latex{a^2+b^2\gt c^2}.
Whereas in the case of an obtuse angle:
\latex{a^2+b^2\lt c^2}.
The following question might arise: can we transform the inequality so that an equality is obtained, which can be applied for a general triangle?
The relation is obtained by applying the dot product of the vectors. Let us consider an arbitrary triangle, and let us direct its sides as shown in Figure 28.
Since \latex{\overrightarrow{AB}=\overrightarrow{CB}-\overrightarrow{CA}}, when squaring this equality we obtain the following:
\latex{\overrightarrow{AB}^2=\left(\overrightarrow{CB}-\overrightarrow{CA}\right)^2=\overrightarrow{CB}^2+\overrightarrow{CA}^2-2\times\overrightarrow{CB}\times\overrightarrow{CA}}.
Based on our notations:
\latex{\overrightarrow{AB}^2=c^2}, \latex{\overrightarrow{CB}^2=a^2}, \latex{\overrightarrow{CA}^2=b^2} and
\latex{\overrightarrow{CB}\times\overrightarrow{CA}=\left|\overrightarrow{CB}\right|\times\left|\overrightarrow{CA}\right|\times\cos{\gamma}=ab\times\cos{\gamma}}.
\latex{\overrightarrow{CB}\times\overrightarrow{CA}=\left|\overrightarrow{CB}\right|\times\left|\overrightarrow{CA}\right|\times\cos{\gamma}=ab\times\cos{\gamma}}.
By substituting the expressions into the equation the following is obtained:
\latex{c^2=a^2+b^2-2ab\times\cos{\gamma}}.
This relation can be obtained similarly for any side of the triangle:
\latex{a^2=b^2+c^2-2bc\times\cos{\alpha}};
\latex{b^2=a^2+c^2-2ac\times\cos{\beta}}.
\latex{b^2=a^2+c^2-2ac\times\cos{\beta}}.
The obtained relation is the cosine rule, according to which:
\latex{\overrightarrow{CA}}
\latex{\overrightarrow{CB}}
\latex{\overrightarrow{AB}}
\latex{\gamma}
\latex{ C }
\latex{ B }
\latex{ A }
\latex{ c }
\latex{ a }
\latex{ b }
Figure 28
THEOREM: The square of the length of one side of a triangle is obtained by subtracting twice the product of the lengths of the other two sides and the cosine of the angle included from the sum of the squares of the lengths of the other two sides.
\latex{c^2=a^2+b^2-2ab\times\cos{\gamma}}
If \latex{\gamma = 90^{\circ}}, i.e. the triangle is right-angled, then because \latex{\cos 90^{\circ} = 0} the cosine rule transforms itself to the Pythagorean theorem. That means the Pythagorean theorem is a special case of the cosine rule.
However if we know for a triangle that \latex{a^2 +b^2 = c^2}, then we obtain the following:
\latex{\begin{array}{rcl}2ab\times\cos{\gamma}=0,\\ \cos{\gamma}=0,\\\gamma=90^{\circ},\end{array}}
so the triangle is right-angled. This train of thought verifies the converse of the Pythagorean theorem, which can be formulated as follows:
THEOREM: If in a triangle the sum of the squares of the lengths of two sides is equal to the square of the length of the third side, then the triangle is right-angled.
Example 1
The lengths of the sides of a triangle are: \latex{a = 2}, \latex{b = 4} and \latex{c = 3}. Let us determine its angles.
Solution
Let us denote the angles of the triangle the usual way (Figure 29). Let us write down the cosine rule for the side a of the triangle:
\latex{a^2=b^2+c^2-2bc\times\cos{\alpha}}.
It implies that for the anglea that \latex{2bc \times \cos{\alpha} = b^2 + c^2 – a^2}, from which:
\latex{\cos{\alpha}=\frac{b^2+c^2-a^2}{2bc}=\frac{4^2+3^2-2^2}{2\times4\times3}=0.875\Rightarrow\alpha\approx28.96^{\circ}}.
We can get another angle of the triangle in a similar way:
\latex{\cos{\beta}=\frac{a^2+c^2-b^2}{2ac}=\frac{2^2+3^2-4^2}{2\times2\times3}=-0.25\Rightarrow\beta\approx104.48^{\circ}}.
Based on the sum of the interior angles the third angle is:
\latex{\gamma=180^{\circ}-28.96^{\circ}-104.48^{\circ}=45.65^{\circ}}.
\latex{\gamma}
\latex{\alpha}
\latex{\beta}
\latex{ B }
\latex{ A }
\latex{ C }
\latex{ a }
\latex{ b }
\latex{ c }
Figure 29
Example 2
The larger hand of the church clock of the dome in Szeged (Hungary) is \latex{ 3\, m } long, and the shorter hand is \latex{ 2\, m } long. How far will the end-points of the hands be at exactly \latex{ 8 } o’clock?
Solution
The angle included between the hands at \latex{ 8 } o'clock is: \latex{4\times\frac{360^{\circ}}{12}=120^{\circ}}.
Based on the cosine rule the distance between the end-points of the hands is:
Based on the cosine rule the distance between the end-points of the hands is:
\latex{d^2=3^2+2^2-2\times2\times3\times\cos{120^{\circ}}=19}, which implies \latex{d\approx 4.36} m.
Example 3
In a quadrilateral \latex{ ABCD } the length of the line segment \latex{AB = 32} and the angles marked in the figure are known: \latex{\alpha = 67^{\circ}},\latex{\beta = 41^{\circ}},\latex{\gamma = 38^{\circ}} and \latex{\delta = 80^{\circ}}. Let us determine the length of the side \latex{ CD }. (Figure 30)
\latex{\alpha}
\latex{\beta}
\latex{\gamma}
\latex{\delta}
\latex{ C }
\latex{ B }
\latex{ A }
\latex{ 32 }
\latex{ D }
Figure 30
Solution
It is practical to determine the sides \latex{ AC } and \latex{ AD } of the triangle \latex{ ACD }, since in view of the angle included between them and by applying the cosine rule the side \latex{ CD } can be calculated.
By using the angles the angle of the triangle \latex{ ABD } at the vertex \latex{ D } is \latex{34^{\circ}}, thus
\latex{AD=\frac{32\times\sin{38^{\circ}}}{\sin{34^{\circ}}}\approx35.23.}
In the triangle \latex{ ABC } the angle at the vertex \latex{ C } is \latex{21^{\circ}}, thus:
\latex{AC=\frac{32\times\sin{118^{\circ}}}{\sin{21^{\circ}}}\approx78.84.}
Then in the triangle \latex{ ACD } according to the cosine rule:
\latex{CD^2=35.23^2+78.84^2-2\times25.23\times78.84\times\cos{67^{\circ}}\approx 5286.36.}
This implies the following:
\latex{CD\approx\sqrt{5286.36}\approx72.7.}
Example 4
The ratio of two sides of a triangle is \latex{3 : 2}, the angle included between them is \latex{60^{\circ}} and its third side is \latex{c = \text{30 cm}}. Give the length of the missing sides and the measure of the missing angles.
Solution
In view of the ratio of the sides let us denote these as follows: \latex{a = 3x,\;\; b = 2x}. Let us write down the cosine rule in terms of the known angle:
\latex{\begin{array}{lcl}c^2=(3x)^2+(2x)^2-2\times3x\times2x\times\cos{60^{\circ}};\\ 30^2=(3x)^2+(2x)^2-2\times3x\times2x\times\frac{1}{2};\\ 900=7x^2.\end{array}}
It implies the following:
\latex{x=\sqrt{\frac{900}{7}}=\frac{30}{\sqrt7}=\frac{30\times\sqrt7}{7}.}
So the lengths of the missing two sides:
\latex{a=3\times\frac{30\times\sqrt7}{7}=\frac{90\times\sqrt7}{7}\approx 34; \;\; b=2\times\frac{30\times\sqrt7}{7}=\frac{60\times\sqrt7}{7}\approx22.67.}
In view of these another angle of the triangle can be determined:
\latex{\cos{\alpha}=\frac{b^2+c^2-a^2}{2bc}=\frac{22.67^{2}+30^2-34^2}{2\times22.67\times30}=0.1896\Rightarrow\alpha\approx79^{\circ}.}
The measure of the third angle of the triangle is:
\latex{\beta=180^{\circ}-60^{\circ}-79^{\circ}=41^{\circ}.}
◆ ◆ ◆
It can be proven that the two trigonometric theorems expressed for triangles are not independent. The cosine rule can be derived from the sine rule and conversely. Because of this it always depends on the features of the exercise and the given data which one to use during the solution.
Simple cases (data)
Theorem
Missing data that can be
calculated
calculated
one side and two angles
two sides and the angle
opposite the longer side
opposite the longer side
two sides and the angle included
between them
between them
three sides
sine
sine
cosine
cosine
missing side
the (acute) angle
opposite the shorter side
opposite the shorter side
the third side
the third side
Exercises
{{exercise_number}}. Two sides of a triangle are \latex{a = \text{10 cm}}, \latex{b = \text{15 cm}}. The angle included between the two sides is \latex{\gamma = 60^{\circ}}. Find the third side and the other two angles.
{{exercise_number}}. The ratio of two sides of a triangle is\latex{ 5 : 4 }, the angle included between them is \latex{120^{\circ}} and its third side is \latex{ 20\, cm } long. Give the length of the missing sides and the size of the missing angles.
{{exercise_number}}. The parallel sides of a trapezium are \latex{ 1 } and \latex{2+\frac{\sqrt3}{3}}. Its legs are \latex{\sqrt2} and \latex{\frac{2\times\sqrt3}{3}} long. Give the angles of the trapezium.
{{exercise_number}}. The lengths of the sides of a convex quadrilateral are \latex{ 14\, cm }, \latex{ 6\, cm }, \latex{ 10\, cm } and \latex{ 12\, cm } respectively. The angle included between the \latex{ 12\, cm } and the \latex{ 14\, cm } long sides is \latex{41.9^{\circ}}. Give the missing angles of the quadrilateral.
{{exercise_number}}. Two planes are taking off at an airport with a 1 minute difference. One of them is flying to the south, the other one is flying to the north-east at \latex{ 600\, km/h }. After how much time will they be \latex{ 200\, km } away from each other?
{{exercise_number}}. The sides of a triangle are \latex{ 13,14 } and \latex{ 15 } units long. Give the radius of the circle the centre of which lies on the longest side of the triangle and the circle touches the other two sides of the triangle.
{{exercise_number}}. Is there a triangle where the sum of squares of any two sides is less than the square of the third side?
{{exercise_number}}. For which triangles is the following true: the sum of squares of any two sides is greater than or equal to double the square of the third side?