The triangle ABC A.B.C is the opposite side a, B and C of angle a, B and C respectively, and the value range of y = 2Sin? B + sin (2B + 6 parts of π) is obtained in equal proportion sequence

The triangle ABC A.B.C is the opposite side a, B and C of angle a, B and C respectively, and the value range of y = 2Sin? B + sin (2B + 6 parts of π) is obtained in equal proportion sequence

B ^ 2 = AC = a ^ 2 + C ^ 2-2accosb, so CoSb = (a ^ 2 + C ^ 2-ac) / 2Ac = 0.5 {A / C + C / A-1}
>=0.5 {2 root sign (A / C * C / a) - 1] = 0.5, 0

In △ ABC, ∠ C is an obtuse angle, a ′ - B ′ = BC, and ∠ a = 2 ∠ B is proved Please use the knowledge of grade two

According to the sine theorem, a 2-sinb ^ 2 = sinbsin (180-a-b) according to the sum difference product formula: left = (Sina + SINB) (Sina SINB) = sin (a + b) sin (a-b) (expand and merge by oneself) right = sinbsin (a + b), that is: sin (a-b) = SINB because a, B are both

In △ ABC ∵ sin (B + a) = 2Sin (B + C)  sinc = 2sina why?

∵B+A+C=π
sin(B+A)=2sin(B+C)
∴sin(B+A)=sin(π-C)=sinC
2sin(B+C)=2sin(π-A)=2sinA
∴sinC=2sinA

In △ ABC, if sin ^ 2A + sin ^ 2B = 2Sin ^ 2C, then the angle c is (pure angle, right angle, acute angle, 60 degrees)

Sine theorem: A ^ 2 + B ^ 2 = = C ^ 2
Second cosine theorem: A ^ 2 + B ^ 2 = = C ^ 2 + 2abcosc = = 2 C ^ 2
2abcosc = = C ^ 2 exclude right angles and obtuse angles
Bring in 60 degrees to get:
AB = = C ^ 2, that is, a = = b = = C; an equilateral triangle

In the angle ABC, if a? + B? - C? < 0, then the angle ABC is a. acute triangle, B. right triangle, C. obtuse triangle Still some don't understand

a²+ b²-c²＜0
c²＞a²+ b²
According to the cosine theorem
c²=a²+ b²-2abcosC
That is, a 2 + B 2 - 2 a bcosc is more than a 2 + B 2
-2abcosC＞0
cosC＜0
90°＜C＜180°
It's an obtuse triangle

In △ ABC, if ∠ a + ∠ B = ∠ C, then △ ABC is______ Triangle

∵∠A+∠B=∠C，∠A+∠B+∠C=180°，
∴2∠C=180°，
The solution is ∠ C = 90 °
So the answer is: right angle

The relationship between a ^ 2 + B ^ 2 and C ^ 2 is that when the triangle ABC is an acute angle triangle and when the triangle is an obtuse angle triangle, the relationship between a ^ 2 + B ^ 2 and C ^ 2 is The three sides of the triangle ABC are a.b.c. what is the relationship between a ^ 2 + B ^ 2 and C ^ 2 when the triangle ABC is an acute angle triangle and when the triangle is an obtuse angle triangle? Why?

Cosine theorem a ^ 2 = B ^ 2 + C ^ 2 - 2 · B · C · cosa
b^2 = a^2 + c^2 - 2·a·c·cosB
c^2 = a^2 + b^2 - 2·a·b·cosC,
Therefore, the acute triangle a ^ 2 + B ^ 2 > C ^ 2
Obtuse triangle a ^ 2 + B ^ 2

In the cosine theorem, a 2 > b 2 + C 2 A is an obtuse angle △ ABC is an obtuse triangle. Is not the sum of any two sides of the triangle greater than the third? Why is it less than

It is true that the sum of the two sides is greater than that of the third,
But this a 2 ＞ B 2 + C 2 is the sum of squares on both sides, which is different from the above,
In other words, if the sum of the two sides is greater than the third, there is no guarantee that the sum of squares of the two sides is greater than the square of the third side!

In △ ABC, the opposite sides of ∠ a, ∠ B and ∠ C are respectively a, B and C. It is proved by cosine theorem that when ∠ C is an obtuse angle, a square + b square is less than C square

COSC = a square + b square - C square / 2Ab < 0
A square + b square - C square < 0
A square + b square < C square

In △ ABC, the lengths of edges a and B are the two roots of equation x2-5x + 2 = 0, C = 60 ° and find the length of edge C

∵ the length of a and B is the two roots of the equation x2-5x + 2 = 0,
∴a+b=5，ab=2，
Therefore, A2 + B2 = (a + b) 2-2ab = 21
In ∵ ABC, C = 60,
∴c2=a2+b2-2abcosC=21-2×2×1
2 = 19=
19．
That is, the length of edge C is
19．