A mathematical problem about cosine theorem, will help ah! Thank you~~ Please take a look for me In △ ABC, the opposite sides of angles a, B and C are a, B and C respectively. If (a ^ 2 + C ^ 2-B ^ 2) * tanb = root 3 * AC, then the size of angle B is calculated Solution: CoSb * 2Ac * tanb = radical 3 * AC CoSb * tanb = radical 3 / 2 SINB = root 3 / 2 Why is the answer only 60?

A mathematical problem about cosine theorem, will help ah! Thank you~~ Please take a look for me In △ ABC, the opposite sides of angles a, B and C are a, B and C respectively. If (a ^ 2 + C ^ 2-B ^ 2) * tanb = root 3 * AC, then the size of angle B is calculated Solution: CoSb * 2Ac * tanb = radical 3 * AC CoSb * tanb = radical 3 / 2 SINB = root 3 / 2 Why is the answer only 60?

The answer is wrong. It's 60 or 120

A simple mathematical problem about cosine theorem
Three sides of triangle a, B, C
The square of a plus the square of B is less than the square of C
Is a sharp angle or obtuse angle with B? (right angle is impossible)
It is also possible that the angle between a and B is uncertain

Obtuse angle, of course
According to the cosine theorem: C ^ 2 = a ^ 2 + B ^ 2-2ab * COSC, we can know that:
CosC=(a^2+b^2-c^2)/(2ab)
Because a ^ 2 + B ^ 2

Mathematical problems on cosine theorem
In △ ABC, if 2sinacosb = sinc, then △ ABC must be ()
A. Right triangle B. isosceles triangle
C. Isosceles right triangle D. equilateral triangle

2cosb = sinc divided by Sina
2 (a ^ 2 + C ^ 2-B ^ 2) divided by 2Ac = C divided by a
It is reduced to a ^ 2 + C ^ 2-B ^ 2 = C ^ 2
a^2=b^2
a=b
So it's an isosceles triangle

Sine theorem and cosine theorem
If a person wants to make a triangle and the length of its three heights is 1 / 13, 1 / 11 and 1 / 5 respectively, then he will ()
A. Can't make a triangle that meets the requirements
B. Make an acute triangle
C. Make a right triangle
D. Make an obtuse triangle
In an acute triangle ABC, the opposite sides of angles a, B and C are a, B and C respectively. Then the value of Tanc / Tana + Tanc / tanb is

The first question is B

Proof: the square sum of two diagonals of a parallelogram is equal to the square sum of four sides

It is known that in parallelogram ABCD, AC and BD are its two diagonals. This paper proves: ac2 + BD2 = AB2 + BC2 + Cd2 + ad2. It is proved that if AE ⊥ BC is at point E and the extension of DF ⊥ BC is at point F, then ⊥ AEB = ⊥ DFC = 90 °.⊥ quadrilateral ABCD is a parallelogram, ⊥ AB = DC, ab ∥ CD, ⊥ Abe = ⊥ DCF, ≌ Abe ≌ DCF, ≌ AE = DF, be = CF Ac2 = AE2 + EC2 = AE2 + (bc-be) 2, BD2 = df2 + BF2 = df2 + (BC + CF) 2 = AE2 + (BC + be) 2, ac2 + BD2 = 2ae2 + 2bc2 + 2be2 = 2 (AE2 + be2) + 2bc2, and AE2 + be2 = AB2, that is ac2 + BD2 = 2 (AB2 + BC2). ∵ AB = CD, ad = BC, ∵ ac2 + BD2 = AB2 + BC2 + Cd2 + ad2

Using cosine theorem to prove that the sum of squares of the diagonal length of a parallelogram is equal to the sum of squares of the length of four sides

Let the side lengths of parallelogram be a and B, the diagonal be C and D, and the two adjacent angles be a, and B have a + B = 180, so cosa + CoSb = 0. Apply cosine theorem C ^ 2 = a ^ 2 + B ^ 2-2ab * cosa (1) d ^ 2 = a ^ 2 + B ^ 2-2ab * CoSb (2) (1) + (2) to get C ^ 2 + D ^ 2 = 2 (a ^ 2 + B ^ 2)

Proof: the square sum of two diagonals of a parallelogram is equal to the square sum of four sides

It is known that in parallelogram ABCD, AC and BD are its two diagonals. This paper proves: ac2 + BD2 = AB2 + BC2 + Cd2 + ad2. It is proved that if AE ⊥ BC is at point E and the extension of DF ⊥ BC is at point F, then ⊥ AEB = ⊥ DFC = 90 °.⊥ quadrilateral ABCD is a parallelogram, ⊥ AB = DC, ab ∥ CD, ⊥ Abe = ⊥ DCF, ≌ Abe ≌ DCF, ≌ AE = DF, be = CF Ac2 = AE2 + EC2 = AE2 + (bc-be) 2, BD2 = df2 + BF2 = df2 + (BC + CF) 2 = AE2 + (BC + be) 2, ac2 + BD2 = 2ae2 + 2bc2 + 2be2 = 2 (AE2 + be2) + 2bc2, and AE2 + be2 = AB2, that is ac2 + BD2 = 2 (AB2 + BC2). ∵ AB = CD, ad = BC, ∵ ac2 + BD2 = AB2 + BC2 + Cd2 + ad2

Using cosine theorem to prove that the sum of squares of two diagonals of parallelogram is equal to the sum of squares of four sides

The parallelogram is divided into two triangles, and the cosine theorem is applied respectively

Using cosine theorem to prove that the sum of squares of parallelogram diagonal is equal to the sum of squares of four sides

Why is the square of AC followed by minus, not plus 2 abcosb

Ask a math problem of solving triangle part in Senior Two
In ABC, sinA:sinB : sinc = 2: √ 6: (√ 3 + 1), then what is the minimum internal angle of the triangle?
Note: in √ 3 + 1, only 3 and 1 under the root sign are added outside the root sign

sinA:sinB If: sinc = 2: root 6: (root 3 + 1), then the minimum internal angle of the triangle is? Solution: A: B: C is known from the sine theorem= sinA:sinB If: sinc = 2: √ 6: (√ 3 + 1), the smallest edge is a = 2x, then B = √ 6x, C = (√ 3 + 1) x, the smallest inner angle of triangle is acosa = (b ^ 2 + C ^ 2-A ^ 2) / 2BC = (6 + 2 √ 3) / [2 √ 6 * (√ 3 + 1