Let ABC be A.B.C and cosa = - 1 / 2, then Tan (B + C-A) =?

Let ABC be A.B.C and cosa = - 1 / 2, then Tan (B + C-A) =?

From cosa = - 1 / 2, we can get ∠ a = 120 degrees
Then, B + C - a = 180-2 and a = - 60 degrees
tan(B+C-A)=tan(-60)=-√3

Let ABC be the opposite sides of the three inner angles ABC of the triangle ABC respectively, and s be the area of the triangle ABC. Given that a = 4, B = 5, s = 5, the root sign 3 is used to find the angle c and the edge C

According to the triangle area formula, s = 1 / 2 * absinc,
So sinc = 2S / (AB) = 10 √ 3 / 20 = √ 3 / 2,
So C = π / 3 or C = 2 π / 3,
According to the cosine theorem, C ^ 2 = a ^ 2 + B ^ 2-2abcosc,
So C = √ (a ^ 2 + B ^ 2-2abcosc),
When C = π / 3, C = √ (16 + 25-2 * 4 * 5 * 1 / 2) = √ 21,
When C = 2 π / 3, C = √ (16 + 25 + 2 * 4 * 5 * 1 / 2) = √ 61

In △ ABC, let the edges opposite angles a, B, and C be a, B, and C respectively. We know that a + C = 2B, and
Sinasinc = the square B of COS, the area of triangle is 4 root sign 3, find three sides a B C
Don't be too detailed. Just give me some ideas,

sinAsinC=ac
AC = 2Ac: a + C-B
At the same time, SINB can be obtained by sin + cos = 1
I don't know if LZ has a brochure of mathematical formula, which contains the area formula of trigonometric function, but I can't remember. Hehe, follow the above steps to find SINB, and the rest is simple

Senior high school mathematics: we know that the three angles a, B, C of △ ABC are opposite to the side a = 5, B = 7, C = 8
It is known that the three angles a, B and C of △ ABC are opposite to the edges a = 5, B = 7 and C = 8
Urgent request!

The cosine theorem is as follows
cosB=(a^2+c^2-b^2)/(2ac)=(25+64-49)/(2*5*8)=40/80=1/2
So, angle B = 60 degrees
S = 1 / 2acsinb = 1 / 2 * 5 * 8 * sin60 = 20 * radical 3 / 2 = 10 radical 3

Given the vector AB = (4,2) and the vector AC (3,4), then the area of △ ABC is
RT.
Why use sin? What's the point of converting to sin?

According to the module formula of vector [module of vector: if a = (x, y), then | a | 2 = a · a = x ^ 2 + y ^ 2, | a | = √ (x ^ 2 + y ^ 2)], we get | ab | = √ (4 ^ 2 + 2 ^ 2) = √ 20 = 2 √ 5, | AC | = √ (3 ^ 2 + 4 ^ 2) = √ 25 = 5. According to the cosine formula of angle θ between vector a and B [cos θ = (x1 * x2 + Y1 * Y2) / ((√ (x1 ^ 2 + Y1 ^ 2) * √