Tuesday, October 20, 2009

The Trigonometry of the Triangle

http://www.codecogs.com/reference/maths/trigonometery/the_trigonometry_of_the_triangle.php
There are a number of equations associated with triangles. Of these, the best known are the Sine and Cos formulae.

The Sine Formula.

Consider the Triangle ABC with its Circumcircle. Draw the diameter BX through B
13108/img_trig_42.jpg
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Angle BAX = 90 degrees and angle AXB = angle ACB

From the diagram it can be seen that c = 2R sin C
(1)
\displaystyle \therefore\;\;\;\;\;\frac{c}{sin\,C}\;=\;2R
Therefore by symmetry:-
(2)
\displaystyle \mathbf{\frac{a}{sin\,A}\;=\;\frac{b}{sin\,B}\;=\;\frac{c}{sin\,C}\;=\;2R}
Obtuse Case If A is obtuse, angle BXC = 180 - A
13108/img_trig_43.jpg
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(3)
\displaystyle a\;=\;2R\,sin\,(180^0\;-\;A)\;=\;2R\,sin\,A
(4)
\displaystyle \mathbf{\frac{a}{sin\,A}\;=\;\frac{b}{sin\,B}\;=\;\frac{c}{sin\,C}\;=\;2R}

The Cosine Formula

ABC is an acute-angled triangle of height h
13108/img_trig_40.jpg
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(5)
\displaystyle Then\;\;\;\;\;\;\;\;CD\;=\;b\;cos\,C
(6)
\displaystyle \therefore\;\;\;\;\;\;BD\;=\;a\;-\;b\;cos\,C
Using Pythagoras:-
(7)
\displaystyle h^2\;=\;b^2\;-\;(b\;cos\,C)^2
(8)
\displaystyle and\;\;\;\;\;\;h^2\;=\;c^2\;-\;(a\;-\;b\;cos\,C)^2
(9)
\displaystyle \therefore\;\;\;\;\;b^2\;-\;b^2\,cos^2\,C\;=\;c^2\;-\;a^2\;-\;b^2\;cos^2\,C\;+\;2ab\;cos\,C
(10)
\displaystyle or\;\;\;\;\;\;\;\mathbf{c^2\;=\;a^2\;+\;b^2\;-\;2ab\;cos\,C}
NOTE This equation can be re-written in terms of either angle A or B If C is obtuse
13108/img_trig_41.jpg
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(11)
\displaystyle CD\;=\;b\;cos\,(180^0\;-\;C)
(12)
\displaystyle BD\;=\;a\;+\;b\;cos\,(180^0\;-\;C)
(13)
\displaystyle \therefore\;\;\;\;\;h^2\;=\;b^2\;-\;b^2\;cos^2\,(180^0\;-\;C)
(14)
\displaystyle and\;\;\;\;\;h^2\;=\;c^2\;-\;\left(a\;+\;b\;cos\,[180^0\;-\;C] \right)^2
(15)
\displaystyle \therefore\;\;\;\;b^2\;-\;b^2\;cos^2(180^0\;-\;C)\;= c^2\;-\;a^2\;-\;b^2\;cos^2\,(180^0\;-\;C)\;-\;2ab\;cos\,(180^0\;-\;C)
(16)
\displaystyle or\;\;\;\;\;c^2\;=\;a^2\;+\;b^2\;+\;2ab\;cos\,(180^0\;-\;C)
(17)
\displaystyle Thus\;\;\;\;\;\mathbf{c^2\;=\;a^2\;+\;b^2\;-\;2ab\;cos\,C}
It should be noted that the same equation can be applied in both cases.

Area Of A Triangle

13108/img_trig_91.jpg
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The area of a Triangle is a half base times height.
(48)
\displaystyle i.e.\;\;\;\;\;\;Area\;=\;\frac{1}{2}\;BC\times AD
(49)
\displaystyle But\;\;\;\;\;AD\;=\;b\;sin\;C
(50)
\displaystyle \therefore\;\;\;\;\;\;\;\mathbf{Area\;=\;\frac{1}{2}\,ab\,sinC\;=\;\frac{1}{2}\,bc\,sinA\;=\;\frac{1}{2}\,ac\,sin\,B}

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