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Consider the triangle below:
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| Photo by Math Principles in Everyday Life |
Can we get the trigonometric functions of the given triangle? If you will use the trigonometric functions for the right triangle, then it is not applicable for the given triangle because the given triangle is acute triangle and not a right triangle. We can make the given triangle into two right triangles by draw a perpendicular line or altitude from point B to the base which is AC. Let h be the altitude of a triangle as shown in the figure:
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| Photo by Math Principles in Everyday Life |
First, let's derive the Sine Law for the given triangle. Since the given triangle has two right triangles, we can now use the trigonometric functions for the right triangle as follows:
if h = h, then
c Sin A = a Sin C
Rearrange the above equation,
If you will do the same procedure by draw a perpendicular line or altitude from points A and C, then the equation is
Next, let's derive for Cosine Law for the given triangle. The altitude of a triangle divides the base into two parts as shown below:
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| Photo by Math Principles in Everyday Life |
Since the given triangle has two right triangles, we can now use the trigonometric functions for the right triangle as follows:
By Pythagorean Theorem,
a2 = h2 + y2
but b = x + y
y = b - x
Therefore,
a2 = h2 + y2
a2 = (c Sin A)2 + (b - x)2
a2 = (c Sin A)2 + (b - c Cos A)2
a2 = c2 Sin2 A + b2 - 2bc Cos A + c2 Cos2 A
a2 = c2 (Sin2 A + Cos2 A) + b2 - 2bc Cos A
but Sin2 A + Cos2 A = 1
a2 = b2 + c2 - 2bc Cos A
If you will do the same procedure by draw a perpendicular line or altitude from points A and C, then the equations are
b2 = a2 + c2 - 2ac Cos B
c2 = a2 + b2 - 2ab Cos C