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Applications of the trigonometricrelations
Using the theorems proven for triangles makes it possible to determine missing data and to justify further statements. The following examples point out these possibilities.
Example 1
The lengths of the sides of a triangle are: \latex{a=2, b=4} and \latex{c=3}. Let us determine the length of the median corresponding to the shortest side.
Solution
In this case the shortest side is the side \latex{a} of the triangle, so we have to determine the length of the median \latex{s_{a}.} Let us use the notation in the figure. (Figure 31)
Let us write down the cosine rule for the triangles \latex{ ABD } and \latex{ ACD }, and let us use that the following holds for the supplementary angle pairs:
Let us write down the cosine rule for the triangles \latex{ ABD } and \latex{ ACD }, and let us use that the following holds for the supplementary angle pairs:
\latex{\cos (180°-\varphi )=-\cos \varphi}.
\latex{c^{2}=s^{2} _{a} +(\frac{a}{2} )^{2} -2s_{a}\times \frac{a}{2}\times \cos \varphi;}
\latex{b^{2}=s^{2}_{a}+(\frac{a}{2} )^{2} -2s_{a} \times \frac{a}{2}\times \cos \varphi.}
\latex{b^{2}=s^{2}_{a}+(\frac{a}{2} )^{2} -2s_{a} \times \frac{a}{2}\times \cos \varphi.}
By simplifying the equations:
\latex{c^{2}= s^{2}_{a}+\frac{a^{2} }{4}-s_{a}\times a\times \cos \varphi;}
\latex{b^{2}=s^{2}_{a}+\frac{a^{2} }{4}+s_{a}\times a\times \cos \varphi.}
\latex{b^{2}=s^{2}_{a}+\frac{a^{2} }{4}+s_{a}\times a\times \cos \varphi.}
Let us add the two equations, then after the rearrangement the length of the median \latex{s_{a}} is obtained:
\latex{c^{2}+b^{2}=2s^{2}_{a}+\frac{a^{2} }{2};}
\latex{s^{2}_{a}=\frac{2c^{2}+2b^{2}-a^{2} }{4};}
\latex{s_{a}=\frac{\sqrt{2c^{2}+2b^{2}-a^{2} } }{2}.}
\latex{s^{2}_{a}=\frac{2c^{2}+2b^{2}-a^{2} }{4};}
\latex{s_{a}=\frac{\sqrt{2c^{2}+2b^{2}-a^{2} } }{2}.}
Based on the data of the example its value is:
\latex{s_{a}=\frac{\sqrt{2\times 3^{2}+2\times 4^{2}-2^{2} } }{2}\approx 3.39.}
WILLEBRORD S NELLIUS (\latex{ 1591–1626 }),
a Dutch engineer worked out a (trigonometric) method, which was based on determining the data of the triangles; by applying it the maps became more precise. (The method is shown in lesson \latex{ 10 }.)
\latex{s_{a}=\frac{\sqrt{2c^{2}+2b^{2}-a^{2} } }{2}}
a Dutch engineer worked out a (trigonometric) method, which was based on determining the data of the triangles; by applying it the maps became more precise. (The method is shown in lesson \latex{ 10 }.)
\latex{s_{a}=\frac{\sqrt{2c^{2}+2b^{2}-a^{2} } }{2}}
\latex{ C }
\latex{ D }
\latex{ B }
\latex{ A }
\latex{ c }
\latex{ a }
\latex{ b }
\latex{ S_a }
Figure 31
The relation relating to the length of the medians corresponding to the other two sides can be given in a very similar way.
\latex{s_{b}=\frac{\sqrt{2a^{2}+2c^{2}-b^{2} } }{2};}
\latex{s_{c}=\frac{\sqrt{2a^{2}+2b^{2}-c^{2} } }{2}.}
Example 2
Let us determine the distance \latex{ AB } from the following data: this distanceis seen from the point \latex{ C } at an angle of \latex{ 30º }. When walking \latex{ 100\, m } on theangle bisector of the angle closer to the distance to measure we get toa point \latex{ D } at which the direction pointing to the point \latex{ A } includes an angle of \latex{ 120º } and the direction pointing to the point \latex{ B } includes an angle of \latex{ 90º } with the way we walked. Let us give the distance \latex{ AB }. (Figure 32)
\latex{ A }
\latex{ B }
\latex{ D }
\latex{ C }
\latex{ 100\,m }
Figure 32
Solution
Since the line segment \latex{ CD } is an angle bisector, by writing down the sine rule in the triangle \latex{ ACD }:
\latex{\frac{AC}{100} = \frac{\sin 120°}{\sin 45°};}
\latex{AC\approx 122.47 m.}
\latex{AC\approx 122.47 m.}
Similarly in the right-angled triangle \latex{ BDC }:
\latex{\cos 15°= \frac{100}{CB};}
\latex{CB\approx 103.53.}
\latex{CB\approx 103.53.}
Thus two sides and the angle included between them became known in the triangle \latex{ ABC }, therefore by applying the cosine rule:
\latex{AB^{2} = 122.47^{2}+103.52^{2}-2\times 122,47\times 103,52\times \cos 30°;}
\latex{AB^{2}\approx 3758.26.}
\latex{AB^{2}\approx 3758.26.}
So the distance \latex{ AB } is obtained as \latex{\sqrt{3758.26} \approx 61.3 m.}
Example 3
We fix the two ends of a \latex{ 122\, cm } long rope at points that are \latex{ 76\, cm } awayfrom each other. Then we strain it at an internal point so that the twolegs of the rope include an angle of \latex{ 64º }. What are the lengths of thetwo parts of the rope? (Figure 33)
\latex{ B }
\latex{ 76\,cm }
\latex{ A }
\latex{ b }
\latex{ a }
\latex{ C }
Figure 33
Solution
Let us use the notations of Figure 33. In the triangle \latex{ ABC } strained by the rope parts
\latex{a+b=122,c=76} and \latex{\gamma =64°}.
Let us write down the cosine rule for the side \latex{ c }:
\latex{76^{2}=a^{2}+b^{2}-2ab\times \cos 64°.}
Let us use that
\latex{a^{2}+b^{2}= (a+b)^{2}-2ab=122^{2}-2ab.}
From here:
\latex{5776=14884-2ab-2ab\times \cos 64°;}
\latex{ab=\frac{14884-5776}{2+2\times \cos 64°} \approx} 3166.
\latex{ab=\frac{14884-5776}{2+2\times \cos 64°} \approx} 3166.
Since \latex{b=122-a,} the following equation is obtained:
\latex{a\times (122-a)=3166;}
\latex{a^{2}-122a+3166=0.}
\latex{a^{2}-122a+3166=0.}
This implies the following:
\latex{a_{1}\approx 84.6 cm} and \latex{a_{2}\approx 37.4 cm} are obtained.
These are also the lengths of the resulting rope parts.
Example 4
In a quadrilateral \latex{a^{2}+c^{2}=b^{2}+d^{2}.} What is the angle included between thediagonals? (Figure 34)
\latex{ C }
\latex{ B }
\latex{ A }
\latex{ D }
\latex{ c }
\latex{ z }
\latex{ b }
\latex{ y }
\latex{ x }
\latex{ v }
\latex{ d }
\latex{ a }
Figure 34
Solution
Let us denote the lengths of the parts resulting on the diagonals as \latex{ x, y, z } and \latex{ v } according to Figure 34; let the angle included between the diagonals be \latex{\varphi} .
Let us write down the cosine rule for the four triangles defined by the diagonals:
Let us write down the cosine rule for the four triangles defined by the diagonals:
\latex{a^{2}=x^{2}+y^{2}-2xy\times \cos \varphi;}
\latex{b^{2}=y^{2}+z^{2}-2yz\times \cos (180°-\varphi);}
\latex{c^{2}=z^{2}+v^{2}-2zv\times \cos \varphi;}
\latex{d^{2}=x^{2}+v^{2}-2xv\times \cos (180°-\varphi).}
\latex{b^{2}=y^{2}+z^{2}-2yz\times \cos (180°-\varphi);}
\latex{c^{2}=z^{2}+v^{2}-2zv\times \cos \varphi;}
\latex{d^{2}=x^{2}+v^{2}-2xv\times \cos (180°-\varphi).}
By using that \latex{\cos (180°-\varphi)=-\cos \varphi,} let us add the squares of the opposite sides:
\latex{a^{2}+c^{2}=b^{2}+d^{2;}}
\latex{x^{2}+y^{2}-2xy\times \cos \varphi+z^{2}+v^{2}-2zv\times \cos \varphi=\\ y^{2}+z^{2}+2yz\times \cos \varphi+x^{2}+v^{2}\times \cos \varphi}
\latex{x^{2}+y^{2}-2xy\times \cos \varphi+z^{2}+v^{2}-2zv\times \cos \varphi=\\ y^{2}+z^{2}+2yz\times \cos \varphi+x^{2}+v^{2}\times \cos \varphi}
After rearranging it so that one side of the equation is equal to \latex{ 0 }:
\latex{2\times (xy+zv+yz+xv)\times \cos \varphi=0}
The left-hand-side can only be equal to \latex{0}, if \latex{ \cos } \latex{\varphi=0}, i.e. \latex{\cos \varphi=90°}.
So in the case of the given conditions the diagonals will be perpendicular to each other.
Note: The converse of the statement mentioned in the example can directly be obtained from the Pythagorean theorems that can be written down. I.e. if the diagonals of a quadrilateral are perpendicular, then the sums of squares of the lengths of the opposite sides are equal.
So in the case of the given conditions the diagonals will be perpendicular to each other.
Note: The converse of the statement mentioned in the example can directly be obtained from the Pythagorean theorems that can be written down. I.e. if the diagonals of a quadrilateral are perpendicular, then the sums of squares of the lengths of the opposite sides are equal.
Exercises
{{exercise_number}}. The sides of a triangle are \latex{ 4\, cm }, \latex{ 5\, cm } and \latex{ 6\, cm } long. Give the length of the medians.
{{exercise_number}}. Two ships leave a port at the same time; the angle included between their directions is \latex{ 100º }.The velocity of one of them is \latex{ 50\, km/h }, the velocity of the other one is \latex{ 60\, km/h }. How faraway will they be from each other in \latex{ 5 } hours?
{{exercise_number}}. The ratio of the angles of a triangle is \latex{2\div 7\div 9}. Give the sides if \latex{a=50 cm}.
{{exercise_number}}. The angle opposite the \latex{ 4\, cm } long side of a triangle is \latex{ 40º }, the median corresponding to theside is \latex{ 5\, cm } long. What are the lengths of the other two sides?
{{exercise_number}}. The \latex{ 4\, cm } long angle bisector of the right angle divides a right-angled triangle with a \latex{ 10\, cm } long hypotenuse into two partial triangles. Calculate the angles and sides of these.
{{exercise_number}}. Lightning bolt is seen at an angle of \latex{ 45º }. There is a \latex{ 10\, s } delay between the lightning andthe start of the thunder, and the thunder can be heard for \latex{ 2\, s }. Give the distance coveredby the lightning if the speed of sound is \latex{ 330\, m/s }.
{{exercise_number}}. The lengths of the sides of an acute triangle are \latex{ a },\latex{ b } and \latex{ c }, its angles are \latex{\alpha ,\beta} and \latex{\gamma} respectively. The altitudes intersect the respective opposite sides at the points \latex{ P }, \latex{ Q } and \latex{ R }. Give the perimeter of the base point triangle \latex{ PQR } in view of the data of the triangle \latex{ ABC }.
{{exercise_number}}. One angle of a triangle is \latex{ 120º }; the length of one of its sides is equal to the arithmetic meanof the lengths of the other two sides. Give the other two angles.
