It is proved that if the radius of the circumscribed circle of a triangle is r, then a = 2rsina, B = 2rsinb, C = 2rsinc How to prove a = 2rsina, B = 2rsinb, C = 2rsinc

It is proved that if the radius of the circumscribed circle of a triangle is r, then a = 2rsina, B = 2rsinb, C = 2rsinc How to prove a = 2rsina, B = 2rsinb, C = 2rsinc

Let the center of circumscribed circle be o, connect Bo and extend the intersection circle to d
Then we can see that in triangle BCD, angle BCD is right angle, BD = 2R, angle BDC = angle a, so a = 2rsina
Similarly, we can get b = 2rsinb, C = 2rsinc

It is proved that if the radius of circumcircle of triangle is r, then a = 2rsina, B = 2rsinb, C = 2rsinc
Write the solution of obtuse triangle,
Right angle and acute angle are not needed

A > 90 ° of triangle
Make the diameter cross circle B, another point to D. connect CD
∠D=180°-∠A,∠DCB=90°
a=BC=BD*sinBDC=2Rsin(180-∠A)=2RsinA

It is proved that if the radius of circumscribed circle of triangle is r, then a = 2rsin a B = 2rsin B C = 2rsin C

We only prove that a = 2rsina, the remainder is the same
Draw a triangle ABC and its circumscribed circle, the center of which is O, connect Co, and extend the intersection circle to D. the angle a corresponding to the arc of BC is equal to the angle BDC. In the right triangle BCD, where the angle B is a right angle (the circumferential angle corresponding to the diameter is a right angle), then Sina = sinbdc = A / 2R

It is known that △ ABC is an equilateral triangle with a side length of 6cm

Make ad ⊥ BC, intersect BC at D, intersect circumscribed circle at e, make BF ⊥ AC, intersect AC at F, intersect ad at O △ ace as right triangle, ∠ CAE = 30 ° AE = AC / cos30 ° = 4 √ 3cm, radius of circumscribed circle = AE / 2 = 2 √ 3cm, radius of inscribed circle = ad-ao = ad-ae / 2 = √ 3cm, radius of circumscribed circle: half of inscribed circle

The side length of an equilateral triangle ABC is a. find its inscribed circle radius and circumscribed circle radius

Inscribed circle
Circumscribed circle -- root of 3

The side length of the regular triangle ABC is 2A. Let the radius of the inscribed circle of the triangle ABC be r and the radius of the circumscribed circle be r

2:1

In the triangle ABC, a = 30 ° and the diameter of the circumscribed circle of the triangle ABC is 2R, the maximum area of the triangle can be obtained

R2 / 4-8 radical 3 (for reference only)

If ABC, ABC and ABC are known, we can find the height of a side?

Sorry, I don't know any formula. Let a side diagonal be angle a, B side diagonal be angle B, C side diagonal be angle C. make ad perpendicular to D (I use "2 for square) ad" 2 = AC "2 + CD" 2 = B "2 + (A-C CoSb)" 2 = B "2 + (a - (a" 2 + C "2-B" 2) / 2a) "2 (cosine theorem CoSb = (a" 2 + C "2-B" 2) / 2Ac)

In the triangle ABC, the length of the middle line on the side BC is m, and M is represented by three sides a, B and C. The formula is_____
There should be a detailed process

In the triangle ABC, the midpoint of BC is D, the midline of BC is ad = m, BD = CD = A / 2, ∠ ADC = π - ∠ ADB, cos ∠ ADC = - cos ∠ ADB. According to the cosine theorem m ^ 2 + A ^ 2 / 4-b ^ 2 = MacOS ∠ ADC = - MacOS ∠ ADB = C ^ 2-m ^ 2-A ^ 2 / 4, 2 * m ^ 2 = B ^ 2 + C ^ 2-A ^ 2 / 2, M = (1 / 2) √ (2 * B ^ 2 + 2 * C ^ 2-A ^ 2)

If the three sides of △ ABC are a, B, C, and its area is A2 + B2 − C24, then the inner angle c equals ()
A. 30°B. 45°C. 60°D. 90°

The area of ∵ △ ABC is s = A2 + B2 − C24, so 4S = A2 + b2-c2, 4 × 12absinc = 2abcosc can be obtained from cosine theorem, sinc = COSC. ∵ 0 < C < π, ∵ C = π 4, so the answer is: π 4