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Mandarinka [93]
3 years ago
15

How do similar right triangles lead to the definitions of the trigonometric ratios?

Mathematics
1 answer:
zheka24 [161]3 years ago
6 0
The ratio of two sides of one right triangle is the same of the corresponding sides of any its similar triangles.

Call a, b, c the sides of a right triangle and A, B, C the sides of any of its similar triangles.

Then a/b  = A/B, which means that this ratio is a constant.

The same for a/c = A/C, and b/c = B/C.

Even, the same is true for the inverses: b/a = B/A; c/b = C/B, and d/a = C/A.

Then, you can define a function for every one of these ratios. Those functions are the trigonometric functions.


 
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Can someone please clearly and correctly solve and show me how to work out QUESTION 3 AND 4??
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There are 2 options to solve that.
1. The first one is by derivatives. 
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you have x so for (-6) solve the first equation, then you find y
y=(-6)^2+12*(-6)+36=(-72)
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for x we have:
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you have x so put x into the main equation
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and we have the same solution: vertex is (-6, -72)

For next task, I will use the second option:
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for that task we have
b=(-6)
a=1
x=(6)/2=3
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a-x

b-/

Step-by-step explanation:

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Step-by-step explanation:

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