Given that the radius of circle O is R and the length of chord AB is also R, find the degree of ∠ AOB The title is like the title. The red flag is yours

Given that the radius of circle O is R and the length of chord AB is also R, find the degree of ∠ AOB The title is like the title. The red flag is yours

If the length of chord AB is R and the radius is r, then AB and two radii can form an equilateral triangle, so the angle AOB is 60 degrees

If the radius of circle O is 2cm, find the length of chord ab

AOB = 360 / (1 + 5) = 60 or = 5 * 360 / (1 + 5) = 300 triangle AOB is an equilateral triangle (angle AOB = 60, OA = ob = R), so AB = r = 2cm

The chord Mn divides the circle O into two arcs with a degree ratio of 4:5. If t is the midpoint of Mn, then ∠ MOT is equal to

The chord Mn divides the circle O into two arcs, and their degree ratio is 4:5, then the angle mon = 360 * 4 / 9 = 160 degrees, and the angle mot = 160 / 2 = 80 degrees

The chord Mn divides the circle O into one to three, connects OM and on, makes AB parallel on through the midpoint a of Mn, intersects the arc Mn with B, and calculates the degree of arc BN

The chord Mn divides the circle O into 1:3
The arc Mn = 90 °, it is easy to know that the straight line AB passes through the mid point of OM, ∠ BOM = 60 °, bon = 30 ° and arc BN = 30 °

The chord Mn divides the circle O into one to three, connects OM, on, and makes AB parallel on through the midpoint a of Mn, and intersects the arc Mn with B. why can we find the degree of arc BN, ∠ BOM = 60 degrees

The BA was extended to cross om at point C and connected with BM,
Because the chord Mn divides the circle O into three
So ∠ mon = 90,
Because BA ‖ on, a is the midpoint of Mn
So AC divides the OM vertically,
So BM = Bo,
Om = ob
So om = ob = BM
So △ mob is an equilateral triangle
So ∠ BOM = 60 °

As shown in the figure, in △ ABC, ∠ ACB = 90 ° and ∠ B = 36 ° with C as the center and Ca as the radius intersects AB at point D and BC at point E AD、 Degree of de

Connecting CD, ∵ △ ABC is a right triangle, ∵ B = 36 °, a = 90 ° - 36 ° = 54 °, ∵ AC = DC,  ADC = ∠ a = 54 °, ∵ ACD = 180 ° - ∠ a - ∠ ADC = 180 ° - 54 ° = 72 °,  BCD = ∠ ACB - ∠ ACD = 90 ° - 72 ° = 18 °, ? ACD and ∠ BCD are ad, de respectively

In △ ABC, ∠ ACB = 25 ° with C as its center and Ca length as its radius intersects AB at d to find the degree of arc ad

The degree of arc ad is 25 degrees
The degree of the arc is equal to the degree of the angle to the center of the circle

The knowledge solution to find the length of AD by the intersection of a circle AB with C as its center and Ca as its radius

Is it right? Answer with the knowledge of cosine
cos∠CAD=5/√﹙12²+5²﹚=5/13   ∠ACD=180-2∠CAD
AD²=5²+5²-2×5×5cos∠ACD
      =50×﹙1-cos∠ACD﹚
     = 50×﹙1+cos2∠CAD﹚
     = 50×﹙1+2cos²∠CAD-1﹚
     =100cos²∠CAD
      =100×5²/13³
∴AD=50/13 

As shown in the figure, in RT △ AOB, ∠ B = 40 ° with OA as the radius and o as the center of the circle as ⊙ o, intersection AB at point C, intersection ob at point D Degree of CD

Connect OC,
∵∠O=90°，∠B=40°，
∴∠A=180°-90°-40°=50°，
∵OA=OC，
∴∠ACO=∠A=50°，
∴∠COD=∠ACO-∠B=10°，
Qi
The degree of CD is 10 degrees

As shown in the figure, in RT △ AOB, ∠ AOB = 90 ° and the circle with OA as radius intersects AB at point C. If Ao = 5 and ob = 12, find the length of BC

Pass through point E as OE ⊥ AC at point E,
∵∠AOB=90°，AO=5，OB=12，
∴AB=13，
∴EO×AB=AO×BO，
∴EO=AO×BO
AB=5×12
13=60
13，
In RT △ AEO
AE=
AO2−EO2=25
13，
∴AC=25
13×2=50
13，
∴BC=13-50
13=119
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