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The sine rule
Reminder
It is true for the sides and the angles of a triangle that there is a longer side opposite a greater angle and there is a greater angle opposite a longer side.
Let us take a triangle \latex{ ABC }, and let us denote its sides and angles in the usual way.
(Figure 23)
(Figure 23)
\latex{ A }
\latex{ B }
\latex{ B }
\latex{ A }
\latex{ c }
\latex{ C }
\latex{ C }
\latex{ m_c }
\latex{ m_c }
\latex{ b }
\latex{ a }
Figure 23
In year \latex{ 10 } we verified that the area of a triangle is: \latex{A=\frac{b\times c\times \sin \alpha }{2}}.
The result is obviously satisfied for any two sides and the angle included. Thus we formulated the following theorem:
THEOREM: The area of a triangle is equal to half of the product of the lengths of two sides and the sine of the angle included.
\latex{A=\frac{b\times c\times \sin \alpha }{2}}
By using this statement the following theorem is obtained, which is called the sine rule.
THEOREM: In a triangle the ratio of the lengths of two sides is equal to the ratio of the sines of the angles opposite these sides.
Proof
Based on the area formula we can write down the following equality:
\latex{A=\frac{b\times c\times \sin \alpha }{2} =\frac{a\times c\times \sin\beta }{2}}.
When multiplying both fractions by \latex{ 2 } and dividing them by \latex{ c } the following is obtained:
\latex{b\times \sin \alpha =a\times \sin \beta}.
This equality can be rewritten in the following form, since the length of none of the sides of the triangle and the sine of none of the angles can be equal to \latex{ 0 }:
\latex{\frac{a}{b} =\frac{\sin \alpha }{\sin \beta }}
With this the statement of the theorem is verified.
The sine rule holds for the ratio of any two sides, therefore its statement can be formulated in the following form too:
\latex{\frac{a}{\sin \alpha } =\frac{b }{\sin \beta } =\frac{c}{\sin \gamma }}.
Example 1
We are looking for the distance of the point \latex{ A } on one bank of the river from the point \latex{ C } on the other bank. The distance measured from the point \latex{ B } on the same bank as the point \latex{ A } is known: \latex{ AB = 300\, m }. We determined the measure of the angles shown in the figure, \latex{ a = 50º } and \latex{ b = 60º }. What is the distance between the points \latex{ A } and \latex{ C } ? (Figure 24)
\latex{\alpha}
\latex{\beta}
\latex{ C }
\latex{ B }
\latex{ A }
Figure 24
Solution
From the figure it can be seen that one side and two angles of a triangle are known, and the aim is to determine the length of another side
For the third angle of the triangle \latex{ABC:\gamma =180°-50°+60°=70°}.
According to the sine rule:
\latex{\frac{AC}{AB} =\frac{\sin \beta }{\sin \gamma }}, i.e. \latex{\frac{AC}{300} =\frac{\sin 60° }{\sin70° }}.
From here the following is obtained for the distance \latex{ AC }:
\latex{AC=\frac{300\times \sin 60°}{\sin 70°}\approx 276.48\;m}
Example 2
We would like to get from the point \latex{ A } to the point \latex{ C }, examined in the previous example, in the shortest possible way. In which direction shall we row if the velocity of our boat is \latex{4\frac{m}{s}} in a lake and the velocity of the flow of the river is \latex{1\frac{m}{s}}?
\latex{\overrightarrow{V} _{boat} }
\latex{50°}
\latex{\overrightarrow{V} _{resultant} }
\latex{\overrightarrow{V} _{river} }
\latex{\phi}
\latex{\phi}
\latex{ C }
\latex{ A }
Figure 25
Solution
The resultant velocity of the boat should point to the direction of the line segment \latex{ AC } (Figure 25).
According to the example we know the length of the component vectors and the angle included between the resultant vector and one of its components. Let us determine the measure of the angle denoted by \latex{\phi} in the figure.
According to the sine rule:
\latex{\frac{\sin \phi }{\sin 50°} =\frac{1}{4} }.
It implies the following:
\latex{\sin \phi =\frac{1}{4}\times \sin 50°\approx 0.1915}
The angle can unambiguously be determined from this, since there is a smaller angle opposite the shorter side in the triangle in question: \latex{\phi =11,04°}.
So we have to row in this direction to reach the target in the shortest way.
Example 3
In a triangle \latex{ a = 3 }, \latex{ b = 4 }, and the angle opposite the side a is \latex{ a = 45º }.
Let us determine the missing side and angles of the triangle.
Solution
Let us denote the sides and angles of the triangle the usual way. (Figure 26)
Let us write down the sine rule:
\latex{\frac{a}{b} =\frac{\sin \alpha }{\sin \beta }}.
\latex{\alpha}
\latex{\beta}
\latex{\gamma}
\latex{ C }
\latex{ a }
\latex{ b }
\latex{ A }
\latex{ B }
\latex{ c }
Figure 26
With the given data:
\latex{\frac{3}{4} =\frac{\sin 45° }{\sin \beta }}.
Because of this:
\latex{\sin \beta =\frac{4\times \sin 45°}{3} =\frac{4\times \frac{\sqrt{2} }{2} }{3} =\frac{2\times \sqrt{2} }{3} \approx 0.943}.
Based on the equality \latex{\sin \beta \approx 0.943} two solutions are possible, since all we can say about the angle \latex{\beta} is that \latex{0\lt \beta \lt 180°}.
The solutions can be the following:
\latex{\beta_{1} \approx 70.53°}, and \latex{\beta_{2} \approx 180°-70.53°=109.47°}.
In the first case an acute triangle, in the second case an obtuse triangle is obtained.
(Figure 27)
(Figure 27)
\latex{\beta_{2}}
\latex{\gamma_{2}}
\latex{\beta_{1}}
\latex{\gamma_{1}}
\latex{45°}
\latex{45°}
\latex{ B_1 }
\latex{ A_1 }
\latex{ c_1 }
\latex{ A_2 }
\latex{ c_2 }
\latex{ B_2 }
\latex{ 3 }
\latex{ C_2 }
\latex{ 4 }
\latex{ C_1 }
\latex{ 3 }
\latex{ 4 }
\latex{ a })
\latex{ b })
Figure 27
We can give the missing data of the triangle in the two cases similarly.
- In the first case the third angle is:
\latex{\gamma _{1} = 180° – 45° – 70.53° = 64.47°}.
Based on the sine rule:
\latex{\frac{c_{1} }{a} =\frac{\sin 64.47°}{\sin 45°}}.
It implies the following:
\latex{c_{1}=\frac{3\times \sin 64.47°}{\sin 45°} \approx 3.83}.
b) In the other case:
\latex{\gamma _{2} = 180° -45° -109.47° = 25.53°}.
The side c2 is:
\latex{c_{2}=\frac{3\times \sin 25.53°}{\sin 45°} \approx 1.83}.
So we have determined the sides and the angles of the triangle.
If we apply the sine rule to determine the angles of a triangle, then based on the relation
\latex{\sin \alpha =\frac{a\times \sin \beta }{b}}
\latex{\sin \alpha} can be determined. But then we have to pay attention that the sine of \latex{\alpha} and \latex{180°-\alpha} are the same. Therefore this equation can have two solutions.
However in any triangle there is always a greater angle opposite a longer side, so in this case if \latex{b \gt a}, then \latex{\beta \gt a}. Then the angle \latex{\alpha} can only be an acute angle.
Example 4
In a triangle \latex{a=3}, \latex{\alpha =30°} and \latex{\beta =70°}. Let us find the area of the triangle.
Solution
Let us write down the sine rule:
\latex{\frac{b}{a} =\frac{\sin \beta }{\sin \alpha }}.
It implies the following:
\latex{b=\frac{a\times \sin \beta }{\sin \alpha } }.
The third angle of the triangle is: \latex{\gamma = 180° – 30° – 70° = 80°}.
Based on the area formula:
\latex{A=\frac{ab\times \sin \gamma }{2} =\frac{a\times \frac{a\times \sin \beta }{\sin \alpha }\times \sin \gamma }{2} =\frac{a_{2}\times \sin \beta \times \sin \gamma }{2\times \sin \alpha }}.
We can determine the area of the triangle in view of one side and the angles:
\latex{A=\frac{3^{2}\times \sin 70°\times \sin 80° }{2\times \sin 30°} \approx 8.33}.
So the area of the triangle is 8.33 area units.
Exercises
{{exercise_number}}. The area of a triangle is \latex{ 40\, cm^2 }. Two adjacent sides are \latex{ 10\, cm } and \latex{ 15\, cm } long. Give the angles and the sides of the triangle.
{{exercise_number}}. The two diagonals of a parallelogram are \latex{ e = 10\, cm }, \latex{ f = 15\, cm }. Give the angle included between the diagonals if the area of the parallelogram is \latex{ 50\, cm^2 }.
{{exercise_number}}. Give the area of the triangle if one of its sides \latex{ a = 10\, cm }, and the measures of the angles on this side are \latex{ 60º } and \latex{ 40º }.
{{exercise_number}}. One side of a parallelogram is \latex{ 15\, cm } long, and one of its diagonals is \latex{ 20\, cm } long. The angle included between the diagonals is 40º. Give the length of its other side and its area.
{{exercise_number}}. We divided one angle of a regular triangle with \latex{ 30\, cm } long sides into three equal parts. Give the ratio of the lengths of the parts resulting on the opposite side when divided by the trisecting straight lines.
{{exercise_number}}. The difference of the lengths of two sides of a triangle is \latex{ 10\, cm }, the sizes of the angles opposite these sides are \latex{ 60º } and \latex{ 50º }. Find the lengths of the sides of the triangle.
{{exercise_number}}. We divided the circumference of a circle with a radius of \latex{ 1\, m } in the circumference ratio \latex{ 2 : 3 : 4 }. We connected the resulting points of the division. Give the perimeter and the area of the triangle obtained.
{{exercise_number}}. The size of one of the angles of a right-angled triangle with a one-unit-long hypotenuse is \latex{ 60º }. We draw a square above the hypotenuse and equilateral triangles above the legs of this triangle so that these all lie outside the triangle. Calculate the area of the quadrilateral determined by the newly obtained four vertices.
