The triangle is a part of plane that is bounded by three line segments \(\overline{\rm AB}\), \(\overline{\rm BC}\) and \(\overline{\rm BC}\) that are called sides of a triangle as shown in the following figure.
The points of the triangle are usually labeled with letters A, B and C and these are vertices of triangles. As seen from previous figure the inner angles are usually labeled as \(\alpha, \beta\) and \(\gamma\). The sum of inner angles of the triangle is equal to 180° and is calculated using formula which can be written in the following form: $$\alpha + \beta + \gamma = 180°. $$
The triangles can be classified by the lengths of sides and by internal angles. Using the lengths of the sides classificiation there are equilateral triangle, isoscales triangle and scalene triangle. The equilateral trinagle has three sides of the same length. This type of triangle is a regular polygon with all angles measuring 60°. The example of equilateral triangle is shonw in the following figure.
The area of the equilateral triangle can be calculated from relation $$ A = \frac{\sqrt{3}}{4} a^2.$$
The perimeter of the equilateral triangle can be written as: $$ p = 3\cdot a. $$
The radius of the circumscribed circle i.e. the circle that passes through all vertices of equilateral triangle can be written as: $$R = \frac{a}{\sqrt{3}}.$$
The radius of the inscribed circle can be calculated as: $$ r = \frac{\sqrt{3}}{6}a.$$
The altitude (height from any side) can be calculated as: $$ h = \frac{\sqrt{3}}{2}a.$$
The isosceles triangle has two sides of equal length and it is shown in following figure. The isosceles triangle has two angles of the same measure which are opposite to the two sides of the same lenghts.
The height of the isosceles triangle can be calculated using formula derived from Pythagorean theorem as $$ h = \sqrt{b^2 - \frac{1}{4}a^2}.$$
The area is calculated using formula which can be written as:
\begin{eqnarray}
A &=& \frac{1}{2}ah\\
A &=& \frac{1}{2}a\sqrt{b^2 - \frac{1}{4}a^2}\\
A &=& \frac{1}{2}a^2 \sqrt{\frac{b^2}{a^2}-\frac{1}{4}}.
\end{eqnarray}
The scalene triangle has all tree triangle sides of different length. The area of scalene triangle can be calculated using formula: $$ P = \frac{a \cdot v_a}{2} = \frac{b\cdot v_b}{2} = \frac{c\cdot v_c }{2},$$ where \(a, b,\) and \(c\) are scalene triangle sides, and \(v_a, v_b,\) and \(v_c\) are heights on the corresponding sides of the triangle. The other formula for calculating area of the scalene triangle can be written as: $$ P = \sqrt{s(s-a)(s-b)(s-c)},$$ where parameter \(s\) can be calculated from $$ s = \frac{O}{2}=\frac{a+b+c}{2}.$$ The area of scalene triangle can be calculated with radius of incircle or circumcircle using fomrula: $$ P = rs = \frac{abc}{4R}$$. The area of scalene triangle can be calculated with the help of trigonometry as: $$ P = \frac{ab\sin\gamma}{2} = \frac{bc\sin\alpha}{2} = \frac{ac\sin\beta}{2},$$ $$ P = \frac{a^2 \sin\beta \sin \gamma}{2\sin\alpha} = \frac{b^2 \sin\alpha \sin\gamma}{2\sin\beta} = \frac{c^2\sin \alpha \sin \beta}{2\sin\gamma},$$ $$ P= 2R^2\sin\alpha\sin\beta\sin\gamma.$$
The triangles can be classified by the internal angles on right triangle, acute trinagle, and obtuse triangle. The right triangle is type of triangle in which one of the internal angles is right angle. The acute triangle is type of triangle in which one of the internal angles is acute angle. The obuste triangle is type of triangle in which one of the angles is obuste angle.
The altitude of a triangle is a straight line through a vertex and perpendicular to the opposite side. The opposite side is called the base of the altitude, and the foot of the altitude is the point where the altitude intersects the base. The altitude can alternatively be defined as the distance between the base and the vertex. In a triangle the total of three altitudes can be drawn that connects the vertex with opposite base. These three altitudes intersect in a single point called the orthocenter. It should be noted that the orthocenter of a triangle is inside the triangle if its acute triangle type (one of the angles is acute). The example of the altitudes and orthocenter is shonwin the following figure.
The line that goes through a vertex and the midpoint of the opposite side and divides the triangle into two equal areas is called median. The three medians intersect in a single point named centroid or geometric barycenter (denoted by G). The centroid of a rigid traingular object is also the center of mass.
The angle bisector of a triangle is a line that passes through a vertex which cuts the corresponding angle in half. In triangle the three angle bisectors intersect in a single point named incenter (denoted by letter I), whcih is the center of the trinagle's incircle. The incircle is the circle that lies inside the triangle and touches all three sides. The radius of the incircle is called inradius.
The stright line that is passing through the midpoint of the triangle side and is perpendicular to it is called perpendicular bisector. In triangle the three perpendicular bisectors meet in a single point named circumcenter (usually denoted by letter O). The circumcenter point is the center of circumcircle which is the cricle passing through all three vertices. The diameter of this circle is named circumdiameter while radius of such circle is called circumradius.
The trigomomentry of right triangle
In right triangles the trigonometric ratios (sine, cosine, tangent and cotangent) are used to find the unknown angles or lengths off unkown sides. Right triangle has the opposite side, adjacent side and hypotenuse. The example of right triangle is shown in following figure. The opposite side is the side opposite to the angle of interest. The adjecent side is a side that is in contact with angle of interestest and the right angle. The hypotenuse is the side opposite to right angle and it is the longest side of right triangle. In right triangles the Pythagorean theorem is valid i.e. \(c^2 = a^2 + b^2.\)
The sine of an angle is defined as the ratio of the length of opposite side to the lenght of hypotenuse. The formula of sine angle can be writte as:
$$\sin \alpha = \frac{\mathrm{opposite} \quad \mathrm{side}}{\mathrm{hypotenuse}} = \frac{a}{c} $$
The cosine of an angle is the ratio of length of the adjacent side to the lengthh of the hypotenuse. The cosine of an angle in case of right triangle can be written as: $$\cos \alpha = \frac{\mathrm{adjacent}\quad\mathrm{side}}{\mathrm{hypotenuse}} = \frac{b}{c}.$$ The tangent of an angle is the ratio of the lenght of the opposite side to the length of the adjacent side. The tangent can be written in the following form: $$ \tan \alpha = \frac{\mathrm{opposite}\quad \mathrm{side}}{\mathrm{adjacent}\quad \mathrm{side}} = \frac{b}{a} = \frac{\sin \alpha}{\cos \alpha}$$ The cotangent is the inverse tangent function. This means that the cotangent of angle \(\alpha\) can be written as: $$ \cot \alpha = \frac{\mathrm{adjacent}\quad \mathrm{side}}{\mathrm{opposite}\mathrm{side}} = \frac{1}{\tan \alpha} = \frac{\cos \alpha}{\sin \alpha}$$
Triangle Examples
Example 1 If the ratio of the triangle angles as \(8:13:15\) determine the larges triangle angle. Solution - \(\alpha:\beta:\gamma = 8:13:15 \Rightarrow \alpha = 8k, \beta = 13k, \gamma = 15k\) The sum of angles in a triangle is 180° from which the angle values can be determined. \begin{eqnarray}\alpha + \beta + \gamma &=& 180°\\\nonumber 8k + 13k + 15k &=& 180°\\\nonumber 36 k &=& 180°/:36 \\\nonumber k &=& 5°\\\nonumber \alpha &=& 8 k = 8 \cdot 5°= 40°\\\nonumber \beta &=& 13k = 13 \cdot 5°= 65°\\\nonumber \gamma &=& 15k = 15\cdot 5° = 75°\end{eqnarray} From calculated values of angles in a triangle it can seen that the largest angle is gamma \(\gamma = 75°\)
Example 2 The painting is hanging on the wall. The man of 2 [m] in height is at the 2[m] distance from the wall. The angle between the eye and the lower painint edge is 45° and the angle between the eye and the upper paininting edge is 60° shown in following figure.
Determine the height of the painitng. Solution - From the figure the following data is extracted. \(|AB|= 2 [m], \Delta ABC = 60°, \Delta ABD = 45°\) The height of the painting is \(x = |AC|-|AD|\) \begin{eqnarray} \tan 60° &=& \frac{|AC|}{|AB|} \Rightarrow |AC|= 2\cdot \tan 60° = 2\sqrt{3} \\\nonumber \tan 45° &=& \frac{|AD|}{|AB|} \Rightarrow |AD|= 2\cdot 45°= 2\cdot 1 = 2 \\\nonumber x &=& |AC|-|AD|= 2\sqrt{3} - 2 = 1.46 \end{eqnarray}
Example 3 The length of triangle sides are 12, 13 and 21. If all sides increase for the same amount, then they form right triangle. Determine the ammount by which the triangle sides should increase. Solution - The triangle sides are \(12+x, 13 + x,\) and \(21 + x\) and used them in Pythagorean theorem. \begin{eqnarray}(12 + x)^2 + (13+x)^2 &=& (21+x)^2\\\nonumber 144 + 24x + x^2 + 169 + 26x + x^2 &=& 441 + 42x + x^2 \\\nonumber x^2 + 8x - 128 &=& 0 \\\nonumber x_{1,2} &=& \frac{-8 \pm \sqrt{64 + 4\cdot 128}}{2}\\\nonumber x_{1,2} &=& \frac{-8 \pm 24}{2}\\\nonumber x_1 &=& \frac{16}{2} = 8\\\nonumber x_2 &=& -\frac{32}{2} = -16\end{eqnarray} The \(x_2 = -16\) is not the solution since the value of x must be greater than 0. So the solution is \(x_1 = 8\) which means that all sides of triangle must be enlarged by the value of 8.
Example 3 In isoscales triangle the length of the base (\(a\)) is 30 cm while the length of altitude is 20 cm. Determine the length of height on triangle sides. The isoscles triangle is shown in following figure.
Solution - \begin{eqnarray} b&=& \sqrt{v_a^2 + \left(\frac{a}{2}\right)^2} = \sqrt{20^2 + 15^2} \\\nonumber b &=& \sqrt{625} = 25 [\mathrm{cm}]\end{eqnarray} The area of the triangle is \begin{eqnarray}\frac{a\cdot v_a}{2} &=& \frac{b \cdot v_b}{2} \\\nonumber a\cdot v_a &=& b \cdot v_b \\\nonumber v_b &=& \frac{a\cdot v_a}{b} \\\nonumber v_b &=& \frac{a\cdot v_a}{b} = \frac{30\cdot 20}{25} = 24 [\mathrm{cm}]\end{eqnarray}
Example 4 From the vertex of right angle triangle \(\Delta ABC\) the median is drawn of length of 25 cm, and the length of projection from shorter leg on hypotenuse is 18 cm. Determine the radius of a circle that is inside the triangle. The triangle is shown in following figure.
Example 5 The perimeter of triangle \(\Delta A_1B_1C_1\) is 45 cm, and the perimeter of similar triangle \(\Delta A_2B_2C_2 \) is 30 cm. Determine the ratio of triangle areas. Solution
\begin{eqnarray}
k &=& \frac{O_1}{O_2} = \frac{45}{30}= \frac{3}{2}\\
k^2 &=& \frac{9}{4} \Rightarrow \frac{P_1}{P_2} = \frac{9}{4}
\end{eqnarray}
Example 6 Two sides of a triangle are \(a = 4\) and \(b=6\) cm, and their opposite angles are in ratio 1:2. Determine the length of third side.
Solution Using sine theorem the following relation is obtained
\begin{eqnarray}
\frac{4}{\sin \alpha} &=& \frac{6}{\sin 2\alpha}\\
4 \cdot 2 \sin \alpha \cos \alpha &=& 6 \sin \alpha/:2 \\
4\sin \alpha \cos \alpha - 3 \sin \alpha &=& 0\\
\sin \alpha (4\cos \alpha - 3) &=& 0 \\
\sin \alpha &=& 0 \Rightarrow \alpha = 0 + k\pi \\
4\cos \alpha - 3 &=& 0 \Rightarrow \cos \alpha = \frac{3}{4} \Rightarrow \alpha = 41°24'35"\\
\end{eqnarray}
From the cosine theorem the following relation is obtained:
\begin{eqnarray}
a^2 &=& b^2 + c^2 -2bc\cos\alpha\\
4^2 &=& 6^2 + c^2 -12 c\cos \alpha\\
16 &=& 36 + c^2 -12 \frac{3}{4}c \\
c^2 -9c +20 &=& 0\\
c_{1,2} &=& \frac{9\pm \sqrt{81 - 80}}{2} = \frac{9 \pm 1}{2} \\
c_1 &=& \frac{10}{2} = 5 \\
c_2 &=& \frac{8}{2} =4
\end{eqnarray}
The second solution is omitted due to the fact that we already have a triangle side with the same length.
Example 7 The areas of two similar triangles are \(104 [\mathrm{cm}^2]\) and \(26 [\mathrm{cm}^2\). The perimeter of the smaller triangle is 38 cm. Determine the perimiter of hte bigger triangle. \(\) Solution the first step is to determine the ratio between areas $$ k^2 = \frac{P_2}{P_1} = \frac{104}{26} = 4$$ $$k = \sqrt{k^2} = \sqrt{4} = 2 $$ The perimeter of the bigger triangle is the perimiter of the smaller triangle mulitplied with \(k\) i.e. $$ O_2 = k\cdot O_1 = 2\cdot 38 = 76 [\mathrm{cm}] $$
Example 8 If in right triangle the a = 15 cm and b is 6 calculate the \(\tan \alpha\) Solution
\begin{eqnarray}
c^2 &=& a^2 + b^2 \\
c &=& \sqrt{15^2 + 6^2}\\
c &=& \sqrt{225+36} = \sqrt{261}\\
c &=& 16.15 [\mathrm{cm}]\\
\tan \alpha &=& \frac{a}{c}\\
\tan \alpha &=& \frac{15}{\sqrt{261}}
\end{eqnarray}
Example 9 Calculate the \(\cos \alpha\) for a triangle shown in following figure.
Example 10 In isoscale triangle shown in following figure determine the value of angle \(\alpha\) Solution -
\begin{eqnarray}
\beta &=& 73°\\
\alpha + 2\beta &=& 180°\\
\alpha &=& 180° - 146°\\
\alpha &=& 34°
\end{eqnarray}
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