how do you prove the angle between a tangent and radius is 90 degrees?

Answer 1

Assume that the tangent is not perpendicular to the radius.
Now there must be a perpendicular from the center of the circle to the tangent. It must intersect the tangent elsewhere (apart from the point on intersection).
Now since a perpendicular is the shortest distance of a point from a line, the perpendicular must have distance < radius of the circle (since line from point of intersection of tangent to center of circle is a radius).
=> the foot of the perpendicular must lie within the circle.
=>the line meets the circle elsewhere and the line is not really a tangent!
Thus there is a contradiction.
Therefore the radius is the shortest distance.
Consequently, it must be perpendicular to the tangent at the point of intersection.

The answer provided by Bill argues based on the symmetry of the problem.
It can be more simple put this way...
Consider the angle between the radius and the right portion of the tangent and the angle between the radius and the left portion of the tangent.
Now imagine viewing the circle bottom-up instead of from top-to-bottom as we normally do.
Now right becomes left and the left portion becomes the right. But the angles must remain the same right?
Therefore both the angles are equal. As they both add upto 180 degrees, each must be 90.

Answer 2

The first answer is correct, but it sounds quite complicated.

A tangent must only (by definition) touch the circumference at one point. This means it doesn't run along the circumference, or cross it.

Imagine drawing a circle and then drawing a tangent that you think just touches the circumference - but you don't know if it is at 90 degrees to the radius

Then magnify the point where the radius touched the tangent at the circumference say tenfold, adjust the tangent until it again just touches the circumference.

Do this again and again. Do you get the feeling that you are balancing the tangent at one point on the end of the radius and that the angle between them must be 90 degrees?

You can keep on doing this forever, finely tuning the angle ever more closely to 90 degrees.

In effect, what you are doing is performing calculus pictorially and in your head, without the maths!

Answer 3

here we will be working with two
circles and using the two top
semi-circles

construction:-
draw a circle with centre O and
diameter XOA-mark a point C
on the arc of the circle
approximately above O
on the diameter at X,draw
another circle with centre X {and
radius around half that of the first
circle}to cut the first circle at T
join A and T and extend AT another
inch or so to R
join TX
we now have a tangent line ATR
to the circle with centre X and
the normal to the tangent TX,
which is the radius of the
circle with centre X

the angle XTA of triangle XTA is
90 degrees {thales' theorem}
hence,the tangent line ATR
is perpendicular to the radius XT

this holds true no matter the
ratios of the two circles,therefore,
all tangents to a circle
are perpendicular to the radius of
the circle

{it is easy to prove thales' theorem-
2(x+y)=180,hence (x+y)
=90 degrees}

i hope that this helps

Answer 4

i visit't clarify because of lack of figure because figure is major in geometry. yet supplying you with hint First draw a circle and all situations given in celebration. Then draw perspective AOB. Now draw a radius to the XY tangent and draw yet another radius to tangent X'Y'. and yet another radius to the AB tangent. Now there'll variety 4 triangles now tutor 2 pairs of two triangles congruent via SSS try. Then perspective subtended via radius at center will be congruent then upload all angles there addition will be one hundred eighty degree and they alter into 2x+2o=one hundred eighty and then we may be able to tutor that angel AOB is ninety degree........... If any problem you may reply me.....