As shown in the figure AB is the diameter of circle O, AC, AD are chords, and ab bisection angle CAD

As shown in the figure AB is the diameter of circle O, AC, AD are chords, and ab bisection angle CAD

As shown in the figure: ∵ AB is the diameter of circle O and ∵ AC and AD are chords of circle O ∵ and diameter AB bisects the angle formed by AC and ad ∵ CAD (known condition) connects CO and do to form two triangles ACO and triangle ADO (as long as it is proved that ∵ two or three triangles are congruent, AC = AD) proof: ∵

If the diameter of ⊙ o is known to be ab = 2 and there are two chords AC = √ 2 and ad = √ 3 passing through point a, then the area of the inner part of the circle clipped by ⊙ CAD is () cm2

You can get it from the question
∠CAB=π/4 ,∠DAB=π/6
S=(π1^2)/2-[(π1^2)/4-(1×1)/2]-[(π1^2)/6+1/2×(√3/2)]=π/12+1/2-√3/4

1. Given that AB is the diameter of dot 0, AC and AD are chords, ab = 2, AC = radical 2, ad = 1, find the degree of ∠ CAD. 2. The radius of circle O is 2cm, 1. Given that AB is the diameter of dot 0, AC and AD are chords, ab = 2, AC = radical 2, ad = 1, calculate the degree of ∠ CAD 2. Given that the radius of the circle O is 2cm and the circumference angle of chord AB is 60 °, then the length of chord AB is The two questions are not the same and need to be answered separately

Using CAD to solve, very quickly, No.1: make a circle with ab as its diameter, then make a circle with a point as AC and ad as radius, connect two chords, and mark the angle dimension
No.2: make a straight line, use this straight line to make an arc with a circumference angle of 60 degrees, and then make a circle with three points of this arc, and then use the SC scaling command, it is OK

As shown in the figure, AB is the diameter of ⊙ o, chord de vertically bisects radius OA, C is perpendicular foot, chord DF intersects radius ob at point P, de = 2 3, ∠ DPA = 45 ° to find the length of Op

Connecting OD, let the radius of ⊙ o be r, ∵ chord de vertical bisection radius OA, ∵ OC = AC = 12R, ∵ de ⊥ AB, AB is diameter, ᙽ DC = CE = 12de = 12 × 23 = 3. In RT △ DCO, from the Pythagorean theorem, OD2 = DC2 + oc2, R2 = (12R) 2 + (3) 2, the solution is: r = 2,

AB is the diameter of circle O, chord De is vertically bisector radius OA, C is perpendicular foot, chord DF and radius ob intersect at point P, connect EF and EO, if de = 2 times root sign 3, ∠ DPA = 45 I don't have enough grades to send them out. There are some in other places. You can check them. If other questions are the same, they are not the same. (1) find the radius of 0 O and (2) find the area of the shadow part in the figure

Connect of
1. Because OC = 1 / 2od, the angle CDO = 30 ° and because CE = radical 3, OC = 1, so r = 2
2. Angle EOF = 2 × angle EDF = 90 ° shadow area = s sector oef-s triangle OEF = 1 / 4 * π * 4-1 / 2 * 2 * 2 = π - 2

AB is the diameter of ⊙ o, the chord de vertically bisects the radius OA ⊙ and C is the perpendicular foot. The chord DF intersects with the radius ob at point P, connecting EF and EO. If de = 4 √ 3, ∠ DPA = 45 ⊙

(2) Connect of
In RT △ DCP,
∵∠DPC=45°,
∴∠D=90°-45°=45°．
∴∠EOF=2∠D=90°．
The s sector OEF = 90 / 360 × π × 22 = π
∵∠EOF=2∠D=90°,OE=OF=2,
∴SRt△OEF=  1/2×OE×OF=2．
ν s shadow = s sector oef-srt △ OEF = π - 2

AB is the diameter of circle O, the chord De is vertically bisecting the radius OA, C is the perpendicular foot, the intersection point P of chord DF and radius ob is p, connecting EF, EO The angle DPA is 45 ° and find the radius of circle O?

In my opinion, there is still a lack of conditions, and the range can only be determined to be greater than 30 ° and less than 60 °
The radius is 2

In ⊙ o, we know that the diameter ab of ⊙ o is 2, the length of chord AC is root 3, and the length of chord ad is root 2. What is the square of DC It is better to give a diagram and a detailed proof process

If we want to divide two cases of C D on the same side of AB, the angle DAC = angle bad - angle bacbad is a right triangle type (radius = 1 ad = root 2) angle bad = 45 radius = 1 AC = root 3. We can easily get the angle BAC = 30, the square of DAC = 15dc = 2 + 3 - 2 root sign 6cos 15 = 5 - 2 root sign 6 root sign 2 / 2 (root 3 / 2 + 1 / 2) = 5 - (3 + radical 3) = 2 -

In ⊙ o, given that the diameter ab of ⊙ o is 2, the length of chord AC is root three, and the length of chord ad is root two, what is the square of CD

Because AB is the diameter, so the angle ADB and angle ACB are right angles AC = radical (3), ab = 2 then angle cab = 30 ° ad = root (2), ab = 2 then angle DAB = 45 ° so angle DAC = 45 - 30 = 15 degrees. According to cosine theorem cos angle DAC = (AD ^ 2 + AC ^ 2 - DC ^ 2) / (2 * ad * AC) can be solved

It is known that AB is the diameter of ⊙ o, AC and AD are the two chords of ⊙ o, and ab = 16, AC = 8, ad = 8 3. Find the degree of ∠ DAC

As shown in Figure 1, connect BC, BD, ∵ AB is the diameter of ⊙ o, ? C = ∵ d = 90 °, sin  ABC = acab = 12,  ABC = 30 °. ? sin  abd = 32,  abd = 60 °, ∵ DAC = ∠ CBD = 30 °; as shown in Fig. 2: connect OC, BD, ∵ OA = OC = AC = 8, ∵ BAC = 60 °, cos