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Can any one prove this equation: m = tanθ ?

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my_books

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Can any one prove this

This is the equation of a slope of a straight line m = tanθ. But can any one prove it .....
 

Can any one prove this

Yes sure y not
slope is the ratio of change of one line (say Y) with respect to the other (say X)
so it should be given as slope=Y/X
Now apply this to a right triangle, with angle theta and base X and vertical Y
then the slope is given as
Slope=m=Y/X=hypotenuse x sin(theta)/hypotenuse x cos(theta)
Thus
m=sin(theta)/cos(theta)=tan(theta)
 

    my_books

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Can any one prove this

i do not think it is quite difficult!
 

    my_books

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Re: Can any one prove this

y=mx is the eqn for a st line
so m=y/x
m=tan8 means
m=sin8/cos8
what is there to prove in it?
you gave an eqation and how can it be proved??
 

    my_books

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Re: Can any one prove this

my_books,
Slope is defined as rise/run (vertical component/horizontal component).
This is also the definition of Tan.
Regards,
Kral
 

    my_books

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Can any one prove this

Yup, the definition for tanθ is same as slope. But be mindful that the angle θ is defined as the angle between the straight line and horizontal line. this is true even when θ is larger than 90 degree.
 

    my_books

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Re: Can any one prove this

aaronhor,
You are correct. However, the equivalence between slope and Tan is still valid. For example, in the 2nd quadrant, Tan is negative, but the run is negative and the reise is positive, so the slope is also negative. This holds true in all three quadrants as long as you are careful with the signs of the rises and runs.
Regards,
Kral.
 

    my_books

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Can any one prove this

slope=y/x=tanQ
and tanQ=perp/base=y/x
 

Re: Can any one prove this

slope,m = (delta y)/(delta x) = tan (theta) ---- at that particular point
 

Re: Can any one prove this

just complete the triangle and make it a right angled one and then solve using the angles
 

Re: Can any one prove this

y=mx

for any point on this line y1=m *x1
so for point A OC=m *OB ==>m=OC/OB

we know cosθ= OB/OA
sinθ=AB/OA=OC/OA

tg θ=sinθ/cosθ =OC/OB=m
 

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