As shown in the figure, in the RT triangle ABC, the angle B = 90 degrees. If the distance between the point O and the triangle three times is equal, what is the angle AOC? For detailed explanation

As shown in the figure, in the RT triangle ABC, the angle B = 90 degrees. If the distance between the point O and the triangle three times is equal, what is the angle AOC? For detailed explanation

The point where the three sides of the triangle are equal is the intersection point of the three corner bisectors of the triangle. The angle OAC + angle OCA = 1 / 2 (angle a + angle c) = 45 degrees, so the angle AOC = 135 degrees

In the triangle ABC, the angle a = 2, the angle B = 80 degrees, then the angle c is equal to

A = 80 ° and B = 40 ° so ∠ C = 180 ° - 80 ° - 40 ° = 60 °

Triangle ABC is the inscribed triangle of circle O. it is known that AB is equal to 6 angles and ACB is equal to 30 degrees

Six
Connect OA, ob
Angle AOB = 2 ∠ ACB = 60 °
So △ AOB is an equilateral triangle
So OA = ob = AB = 6
O(∩_ Hope to adopt it

If the triangle ABC is inscribed in the circle O, the angle B = 30 degrees, AC = 2, then the radius length of the circle O is?

Using the sine theorem
AC / sin30 degree = 2R
R is the radius, r = 2

As shown in the figure, given the triangle ABC, angle B = 30 degrees, angle c = 45 degrees, ab = 8, find: 2) radius of circle o

Why are you the same as my paper? I'm writing this question and I can't write it out
This is what I found online
Do ad ⊥ BC through point O, connect AO and Co
∵AB=8,∠B=30
So ad = 4
∵∠C=45
∴CD=4
ν AC = 4 root numbers 2
Because it's the same arc AC
∴∠AOC=60°
So △ AOC is an equilateral triangle
The radius is 4 pieces of sign 2

The triangle ABC is inscribed on the circle O, D is on the extension line of radius ob, the angle BCD = angle a = 30 degrees, proving that CD is tangent to the circle o

Connect OC
According to the circular angle theorem, it can be seen that ∠ BOC = 2 ∠ a = 60 °
∵OB=OC,∠BOC=60°
ν Δ OBC is an equilateral triangle
∴∠OCB=60°
∴∠OCD=∠OCB+∠BCD=60°+30°=90°
∴OC⊥CD
The CD is tangent to the circle o

As shown in the figure, in triangle ABC, angle a = angle ACB, CD is bisector of angle ACB, CE is perpendicular to e. it is proved that angle cob = 3 angle DBC Finally, we try to prove that the angle CDB = 3 angle DCB

It seems that there is no need for "CE is perpendicular to e"!
 a = ∠ ACB, CD is the bisector of  DCA = ∠ DCB = 1 / 2  a ? BDC is the external angle of ? ADC ? BDC = ∠ a + ∠ DCA, i.e. ? BDC = 3 ∠ DCB
This question is not difficult. Think about it carefully!

It is known that ad is the height of △ ABC, ∠ bad = 70 ゜, ∠ CAD = 20 ゜. Find the degree of ∠ BAC

① As shown in Figure 1, when high ad is inside △ ABC,
∠BAC=∠BAD+∠CAD=70°+20°=90°；
② As shown in Figure 2, when high ad is outside △ ABC,
∠BAC=∠BAD-∠CAD=70°-20°=50°，
To sum up, the degree of ∠ BAC is 90 ° or 50 °

As shown in the figure, ad and CE are two heights of △ ABC, known as ad = 10, CE = 9, ab = 12 (1) Find the area of △ ABC; (2) Find the length of BC

（1）∵CE=9，AB=12，
The area of △ ABC = 1
2×12×9=54；
(2) Area of △ ABC = 1
2BC•AD=54，
That is 1
2BC•10=54，
BC = 54
5．

Given that a, B, C are sequential three points on ⊙ O and ∠ AOC = 150 °, then the degree of ∠ ABC is______ .

When a, B and C are as shown in Figure 1, connect AB and BC, ∵ AOC and  ABC are the center angles and peripheral angles of the same arc, and ? AOC = 12 × 150 ° = 75 °; when points a, B and C are shown in Figure 2, connect AB and BC, and make AC pair of circular angles ? ADC, ? AOC and ∠ ADC are opposite to the same arc