Application of sine and cosine theorem In the triangle ABC, 2b = a + C, the perimeter is 20, the area is 10 pieces of 3, find the length of three sides,

According to 2B = a + Ca + C + B = 180, 3b = 180, B = 60 degree, area = 0.5acsin, B = 10 roots, 3aC = 40, according to cosine theorem, CoSb = (a * a + C * C-B * b) / 2Ac, B = 20-a-c is brought in, a ^ 2 + C ^ 2 - (20-a-c) ^ = 4040a + 40C = 520A + C = 13, combined with AC = 40, a = 5, C = 8 or a = 8, C = 5, so B = 20-a-c = 7

Cosine theorem: A ^ 2 + B ^ 2-2 * a * b * COSC = C ^ 2 A ^ 2 + C ^ 2-2 * a * c * CoSb = B ^ 2 B ^ 2 + C ^ 2-2 * b * c * cosa = a ^ 2

a^2+b^2-2*a*b*cosC=c^2
a^2+c^2-2*a*c*cosB=b^2
b^2+c^2-2*b*c*cosA=a^2
Your problem should be that a, B and C are the three internal angles of a triangle. Only in this way can we convert them, right
Then add the three equations, a = 180 degrees - b-c
So cosa = cos (180-b-c)
cosA=cos(B+C)
Then, continue to simplify. It's too much trouble. Do it yourself

Cosine theorem, mathematical problem a ^ 2 + C ^ 2 = B ^ 2 + AC to find ∠ C
In △ ABC, a ^ 2 + C ^ 2 = B ^ 2 + AC, and a / C = radical 3 + 1 / 2, find the size of ∠ C

According to the cosine theorem, B & # 178; = A & # 178; + C & # 178; - 2Ac * CoSb and the known condition a & # 178; + C & # 178; = B & # 178; + AC, CoSb = 1 / 2, B = 60 ° so a + C = 120 °, a = 120 ° - C. according to the sine theorem, a / C = Sina / sinc = sin (120 ° - C) / sinc = (sin120 ° cosc-cos120 ° sinc)

What's cosa, CoSb, COSC in cosine theorem,
AC = √ 3 AB = 5 ∠ a = 30 ° then △ ABC is what triangle. But the answer is obtuse angle. With cosine theorem, there are always two unknowns that can't be solved. I don't know how much cosa is

∠A=30°
cosA=√3/2
It should be OK this time

Using sine theorem or cosine theorem to prove a = b * COSC + C * CoSb
Please be more detailed, step by step, thank you very much
It is required to use sine and cosine theorem to prove it without drawing

Major problems of sine and cosine theorem in Senior High School
Generally, I will give you a triangle angle and other relations, and then the first question is how to start with the angle degree or side length. I'm a little confused

You need to see what conditions are given to you. Generally, if you are given an angle, you must use the cosine theorem. Of course, the sine theorem also needs to be used. It seems unclear to say so. It depends on the specific analysis of the topic

What is an obtuse triangle and what is an acute triangle

Obtuse triangle
Definition: a triangle with an obtuse angle is an obtuse triangle
characteristic:
1. The obtuse angle is greater than 90 degrees and less than 180 degrees
2. In an obtuse triangle, the sum of two acute angles is less than the obtuse angle
Acute triangle
Definition: a triangle with three acute angles is called an acute triangle
Property acute angle three angles in a triangle are acute angles

What is an obtuse triangle

Obtuse triangle
Definition: a triangle with an obtuse angle is an obtuse triangle
characteristic
1. The two heights of the obtuse triangle are outside the obtuse triangle, and the other is inside the triangle
2. The obtuse angle is greater than 90 degrees and less than 180 degrees
3. In obtuse triangle, making height is a common auxiliary line
4. In an obtuse triangle, the sum of two acute angles is less than the obtuse angle
5. The sum of internal angles is 180 degrees, and the sum of external angles is 360 degrees

Proof: let the radius of the circumscribed circle of a triangle be r, then a = 2rsina, B = 2rsinb, C = 2rsinc

First, acute angle
Connect ob, OC, because make OD ⊥ BC at point D, because ob = OC, so △ BOC is isosceles triangle, the high bisection of BOC on the side of BC, that is: ⊥ BOC = 2 ⊥ BOD. Point d bisection of BC, that is: BD = (1 / 2) BC = (1 / 2) a --- (1)
So BD = rsin ∠ BOD -------- ②
From the center angle of the circle = 2 times the circumference angle, we can get ∠ BOC = 2 ∠ a, so ∠ BOD = ∠ a ----- ③
① (1 / 2) a = rsina, that is, a = 2rsina
The same can be proved: B = 2rsinb, C = 2rsinc
Right angles are simpler
If AC is hypotenuse, then AC = 2R, ∠ B = 90 degree
a=ACsinA=2RsinA,
c=ACsinC=2RsinC,
b=2R=2RsinB
It can be proved that the other edges are hypotenuse
Obtuse angle:
Suppose ∠ B is an obtuse angle, a = 2rsina, C = 2rsinc, the proof of which is the same as that of an acute triangle, and the following is omitted
On the proof of B = 2rsinb:
It also makes om ⊥ AC from point O to point M. the circumference angle of edge AC = π - ∠ B = (1 / 2) ∠ AOC
So ∠ com = (1 / 2) ∠ AOC = π - ∠ B
(1/2)b=CM=Rsinπ-∠B=RsinB
So B = 2rsinb
It can be proved that other angles are obtuse angles

It is proved that if the radius of circumcircle of triangle is r, then a = 2rsina, B = 2rsinb, C = 2rsinc
But there is one step I don't quite understand
o(∩_ ∩)o...
Draw the triangle ABC and its circumcircle o
Connect OC, ob, make BC, midpoint D, connect OD
Because the angle cob is equal to two angles a
So angle DOB is equal to angle A
So BD equals R * Sina
So BC (a) = 2R * Sina
The other three sides are the same

BD equals R * Sina
Bo extends the intersection point P, BP is the diameter
BP=2R
The BCP is RT ∠
Circle angle on BC chord: ∠ BPC = ∠ a
So: BC = BP * Sina = 2rsina
D is the midpoint of BC
So: BD = BC / 2 = rsina

Application of sine and cosine theorem In the triangle ABC, 2b = a + C, the perimeter is 20, the area is 10 pieces of 3, find the length of three sides,