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Circle geo proof (1 Viewer)

Joshmosh2

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The question is to prove "The angle in a semi-circle is a right angle".

Initially, I thought of the locus of a circle, with x-coordinate being acos(theta) and the y-coordinate being asin(theta) (for convenience, theta will be z)

so x = acosz and y = asinz
and cosz = x/a and sinz = y/a
By using the fact that cos^2z + sin^2 z = 1,
(x/a)^2 + (y/a) ^2 = 1
so x^2 + y^2 = a^2
center (0,0), with radius a units.
This is how far I got. I really don't know what to do next. These steps might not even be necessary..

Well anyways, I thought of a semi circle with point P (acosz,asinz).
Gradients don't seem to help, but maybe pythagoras' theorem?

I feel so close to solving it.. any help please? :blink2:


EDIT: It is probably easier to prove using triangles equal radii, but using what I did, can you prove it from there?
 

Kurosaki

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The question is to prove "The angle in a semi-circle is a right angle".

Initially, I thought of the locus of a circle, with x-coordinate being acos(theta) and the y-coordinate being asin(theta) (for convenience, theta will be z)

so x = acosz and y = asinz
and cosz = x/a and sinz = y/a
By using the fact that cos^2z + sin^2 z = 1,
(x/a)^2 + (y/a) ^2 = 1
so x^2 + y^2 = a^2
center (0,0), with radius a units.
This is how far I got. I really don't know what to do next. These steps might not even be necessary..

Well anyways, I thought of a semi circle with point P (acosz,asinz).
Gradients don't seem to help, but maybe pythagoras' theorem?

I feel so close to solving it.. any help please? :blink2:


EDIT: It is probably easier to prove using triangles equal radii, but using what I did, can you prove it from there?
You can just prove that the angle at the centre is twice the angle at the circumference subtended by the same arc using Euclidean geometry.
Then, when the angle at the centre is pi radians, it goes without saying therefore, that the angle in the semicircle must be 0.5 pi radians.
 

Kurosaki

True Fail Kid
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2014
The question is to prove "The angle in a semi-circle is a right angle".

Initially, I thought of the locus of a circle, with x-coordinate being acos(theta) and the y-coordinate being asin(theta) (for convenience, theta will be z)

so x = acosz and y = asinz
and cosz = x/a and sinz = y/a
By using the fact that cos^2z + sin^2 z = 1,
(x/a)^2 + (y/a) ^2 = 1
so x^2 + y^2 = a^2
center (0,0), with radius a units.
This is how far I got. I really don't know what to do next. These steps might not even be necessary..

Well anyways, I thought of a semi circle with point P (acosz,asinz).
Gradients don't seem to help, but maybe pythagoras' theorem?

I feel so close to solving it.. any help please? :blink2:


EDIT: It is probably easier to prove using triangles equal radii, but using what I did, can you prove it from there?
If you have your heart set on coordinate geometry however:
Let P be the point (x,y)
A(-a,0), B(a,0).


Therefore AP is perpendicular to BP and an angle in a semicircle is a right angle.
 

Joshmosh2

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If you have your heart set on coordinate geometry however:
Let P be the point (x,y)
A(-a,0), B(a,0).


Therefore AP is perpendicular to BP and an angle in a semicircle is a right angle.
Cool, that's similar to how I worked it out, but instead of P(x,y), I used P(acosx,asinx). Still equates to the same answer tho.
 

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