At the angle of 3cm, BCC is equal to 3cm in diameter Find the length of line ad

At the angle of 3cm, BCC is equal to 3cm in diameter Find the length of line ad

∵ the angle c is equal to 90 °, AC is equal to 3 cm, and BC is equal to 4 cm
∴AB=5
∵ BC is the diameter,
Ψ BDC = 90 ° i.e. CD ⊥ AB,
∵S△ABC=AC*BC/2=AB*CD/2,
∴CD=AB*BC/AB=12/5 cm

As shown in the figure, △ ABC is inscribed in ⊙ o, ∠ BAC = 120 ° and ab = AC = 4. If BD is the diameter of ⊙ o, then BD=______ .

∵∠BAC=120°，AB=AC=4，
∴∠C=30°，
∴∠BOA=60°．
And ∵ OA = ob,
The △ AOB is an equilateral triangle
∴OB=AB=4，
∴BD=8．

If the isosceles triangle ABC inscribes a circle, ab = AC, the angle BAC is equal to 120, and ab = 4, how to find the diameter of the circle in detail?

Connecting Ao Bo co angle Bao = 60 degrees = angle ABO = angle boa
That is, the triangle ABO is an equilateral triangle
So Ao = Bo = AB = 4cm
So it's eight centimeters in diameter

The triangle ABC is inscribed in the circle O, the angle BAC = 120 degrees, ab = AC = 4, BD is the diameter of circle O, and the length of BD is calculated

AB = AC = 4 is isosceles triangle
When AE is perpendicular to BC and E, the angle BAE = angle EAC = 60 degrees
If the triangle ABC is inscribed in the circle O, then the AE extension line passes through the circle center O, and BD is a diameter, then OA = ob angle BAE = 60 degrees
Then the triangle boa is an equilateral triangle BD = 2 * ob = 2 * AB = 8

As shown in the figure, the triangle ABC is inscribed in the circle O, the angle BAC = 120 degrees, ab = AC, BD is the diameter of circle O, ad = 6, and the length of DC is obtained

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In triangle ABC, if angle a minus 2 times angle B equals 70, 2 times angle c minus angle B equals 10, then what is angle c? To calculate the process, the answer is 20 degrees

A-2B=70
Then a = 70 + 2B
2C-B=10
Then B = 2c-10
So a = 4C + 50
So 4C + 50 + 2c-10 + C = 180 degrees
C = 20 degrees

Given the triangle ABC, the angle B is twice the angle a, and the angle c is 20 degrees larger than the angle a, then the angle a is equal to () and the options are: a.40 degrees Given the triangle ABC, the angle B is twice the angle a, and the angle c is 20 degrees larger than the angle a, then the angle a is equal to () and the options are: a.40 ° b.60 ° c.80 ° d.90 °

A

In the triangle ABC, if the angle B-angle a-angle C = 20 degrees, what degree is the angle B equal to? .

The angle B is 100 degrees, which can be done by a quadratic equation

As shown in the figure, the three vertices of the triangle ABC are on the circle O, and the bisector of the outer corner of the angle ACB intersects the circle O in E and ef ⊥ BD in F When the shape of triangle ABC changes, does the value of (BF + CF) △ AC change? If the value is not changed, the range of change is requested.

Connect be and CE, make em perpendicular to point M
Then prove that △ AEM and △ bef are congruent
This leads to the conclusion that AF = BF
So BF + CF = am + cm
So (BF + CF) / AC = 1, it remains unchanged

If the area of the inscribed triangle of radius 1 is 1 / 4, if the three sides of the triangle are a, B, C, then ABC=

Because according to the sine theorem: A / Sina = B / SINB = C / sinc = 2R = 2, r = 1 is the radius of the circumscribed circle
So ABC = 8sinasinb sinc
Because according to the area formula s = 1 / 2 * absinc = 1 / 2 * bcsina = 1 / 2 * acsinb = 1 / 4
So ABC = 8sinasinbsinc = 1 / (ABC) ^ 2
That is, ABC = 1