Middle School Mathematics Tangent Problems, Middle School Mathematics Tangent Definition

1) Proof that because ab is the diameter of the circle O, the angle aeb = 90 degrees, so the angle aed + angle bec = 90 degrees, because de cuts the circle o to e, so the angle aed = angle abe, because ce=cb, so the angle bec = angle ebc, so the angle abe + angle ebc = 90 degrees, that is: the angle abc = 90 degrees, and ab is the diameter of the circle o, so bc is the tangent of the circle o.

2) Solution: Because ad, bc are both tangents of the circle O, and ab is the diameter of the circle O, so ad bc, the angle ade + the angle bce = 180 degrees, because ad, de, bc are all tangents of the circle o, so od bisects the angle ade, oc bisects the angle bce, so the angle ado + angle bco = 90 degrees, because the angle ado + angle aod = 90 degrees, so the angle bco = angle aod, the same way angle boc = angle aod, so the triangle aod is similar to the triangle boc, So ad ob=oa bc, because diameter ab=2 root number 5, so radius oa=ob=root number 5, because ad=2, so 2 root number 5 = root number 5 bc

bc=5/2.

Because: de is the tangent of the circle o;

So: angular ceb = angular bae

By because: bc=ce

Angular CBE= Angular BEC

So there is: angular cbe = angular bae

Because angle bae + angle abe = angle aeb = 90 degrees.

So the angle abe + angle cbe = 90 degrees.

So: BC is the tangent of the circle O.

Because CB=CE, the angle CBE=Angle CEB=90°-Angle AED=90°-Angle DAE

So angular CBA+angular bad=180°.

We know that tangent is perpendicular, so the angle abc = angle bad = 90°, so tangent.

According to the tangent length theorem, de=ad=2

Let bc=x=ce

In the 8-character model, cg=x

That's all I can do, and I won't do the rest, I'm sorry.

Connect od to H

The triangle AOH is similar to the triangle ABG.

oh = two-thirds times the root ten.

bg = four tens.

Geometrically, a tangent is a straight line that touches a point on a curve. More precisely, when the cutting line passes through a point on the curve (i.e., the tangent point), the direction of the cutting line is the same as the direction of that point on the curve. In plane geometry, a straight silver line that has only one common intersection with a circle is called a tangent of a circle.

Nature theoremThe tangent of a circle is perpendicular to the radius of its tangent point; A straight line that passes through the non-centered end of the radius and is perpendicular to that radius is a tangent of the circle.

Determination theoremIf a straight line has an intersection point with a circle, and the line connecting the intersection point to the center of the circle is perpendicular to the line, then the line is a tangent of the circle.

Generally available: 1. Vertical evidence radius.

2. Vertical radius testimonial.

Tangent square stroke: For example, y=x 2, use the derivative to find the tangent equation for (2,3) points. Crisper sets the tangent point (m, n), where n=m2, by y'=2x, the tangent slope k=2m.

Tangent equation: y-n=2m(x-m), y-m 2=2mx-2m 2, y=2mx-m 2, because the tangent passes through the point (2, hand 3), so 3=2m*2-m 2, m 2-4m+3=0, m=1 or m=3.

This question is not difficult, all four of the above are true.

According to the equation of the straight line, the coordinates of the two points of c and p can be obtained, so the length of cp, cd and dp can be obtained, and if the nucleus theorem is satisfied, it can be verified that cd is perpendicular to cp, that is, cp is the tangent of the circle.

There must be 2 points e, the absolute value of the ordinate of point e is equal to 4od=4 point e and on a straight line, and the coordinates of 2 e points can be obtained by bringing in the equation of the straight line Zhenqiaoxian.

(1) OA = OD (radius).

Angle A = Angle ODA = 30°

AB = BC angle C = angle A = 30°

In the right triangle cde of de bc, cde=60°

ode=180°-∠oda -∠cde=180° -30° -60°= 90°

od dede is the tangent of the circle o.

2) Right triangle CDB, cd = 3, c = 30° db = bc 2

bc²-db²=bc²-bc²/4=cd²=33bc²/4=3

bc=2ab=bc=2

od=ab/2=1

Right triangle cde.

de=cd/2=√3/2

Right triangle ode.

oe²=de²+od²=3/4+1=7/4oe=√7/2

3) The value range of r is | oe-r |1, i.e. oe-r >1, or r-oe>1

That is, 7 2 -r >1, r< 7 2-1, or r- 7 2>1, r> 7 2+1

r< 7 2-1, or r> 7 2+1

Connecting of, by the question, oaf= ofa, so oaf+ b= ofa+ b=180°- c=90°, so ofe=180°- afo- bfe=90°, i.e. offe, so ef is the tangent of o.

The center of the circle is O, and the ob bc cd is the three tangents of the circle, so the angle ebo=angle fbo angle fco=angle gco be=bf cf=cg cf vertical bc, cf = rab cd, angle abc + angle bcd = 180 °, so the angle fbo + angle fco = 90 °, that is, the angle boc = 90 ° is simple to find the radius of o = of=24 5, be+cg= bf + fc=bc=10

Solution: b = ODB (isosceles triangle).

doc = 2 b (outer angle = not adjacent inner angle and) c = b (isosceles).

again c + doc = 90°

So c = b = 30°