In triangle ABC, if a = 1, B = root 3, a + C = 2B, solve the triangle

In triangle ABC, if a = 1, B = root 3, a + C = 2B, solve the triangle

A+C=180-B=2B
B = 60 degrees
cosB=(a ²+ c ²- b ²)/ 2ac=1/2
c ²- 2=c
So C = 2
Then a ²+ b ²= c ²
So C = 90 degrees
So a = 30 degrees

In the triangle ABC, 2b = a + C, and C / a = (root 3 + 1) / 2, find a, B, C

The sum of the internal angles of the triangle is 180
B=60
A+B=120
Then sinc = sin (120-a). (1)
=√3/2*cosA+1/2*sinA
And C / a = (root 3 + 1) / 2 change it with the sine theorem
It can be seen that 2sinc = (1 + √ 3) Sina. (2)
Combining equations (1), (2) and simplifying
Tana = 1, so
A=45
B=60
C=75

Let the side lengths of the interior angles a, B and C of the triangle ABC be a, B and C respectively, and (2B root 3C) cosa = root 3acosc ① , find the size of angle A ② , if the angle B = π / 6, the length of the center line am on the BC side is root 7, find the area of ▲ ABC

① Make be vertical AC over B and cross AC to e, (2B root 3C) cosa = root 3acosc, so 2B • cosa root 3C • cosa = root 3acosc promotes 2B • cosa = root 3 • CE + root 3 • AE = root 3 • AC = root 3 • B, so cosa = root 3 / 2, a = 30 degrees 2

In triangle ABC, a, B and C are the side lengths of angles a, B and C respectively, a = 2, radical 3, Tan (a + b) / 2 + Tanc / 2 = 4, sinbsinc = cos ^ 2 (A / 2) Find angles a, B, and B, C online! Hurry!

What is the final question?
tan[（A+B）/2]+tanC/2=4,tan(A+B)=tan(π-C）
tan[（A+B）/2]+tanC/2=tan[π/2-C/2]+tanC/2=cot(C/2)+tan(C/2)
=Cos (C / 2) / sin (C / 2) + sin (C / 2) / cos (C / 2) = [sin (C / 2) ^ 2 + cos (C / 2) ^ 2] / (sinc / 2) (COSC / 2) = 2 / sinc = 4, C = π / 6 or 5 π / 6
cos(A/2)=cos[π-(B+C)]/2=sin[(B+C)/2]
cos(A/2)^2=sin[(B+C)/2]^2=[1-cos(B+C)}/2=sinBsinC,1-cosBcosC+sinBsinC=2sinBsinC
Cos (B-C) = 1, B-C = 0 (π does not meet the meaning of the question)
B = C = π / 6, 5 π / 6 shall be rounded off (there shall be no two obtuse angles),

(1 / 2) in triangle ABC, a, B and C are the opposite sides of angles a, B and C respectively. The root number 3 * B / SINB = A / cosa (1) find the size of angle a (2) if (1 / 2) in triangle ABC, a, B and C are the opposite sides of angles a, B and C respectively, and the root number 3 * B / SINB = A / cosa is known (1) Find the size of angle A (2) If B = 1,

1、3b/sinB=a/cosA
By sine theorem: √ 3sinb / SINB = Sina / cosa
Namely: Tana = √ 3;
So: a = π / 3

In acute triangle ABC, cosa = root 5 / 5, SINB = 3 times root 10 / 10, find the angle C. let AB = root 2, find the area of triangle ABC

sin ² A+cos ² A=1
So Sina = 2 √ 5 / 5
Similarly
cosB=√10/10
cosC=cos[180-(A+B)]=-cos(A+B)
=sinAsinB-cosAcosB=√2/2
C = 45 equals
c/sinC=a/sinA
c=2
So a = 4 √ 10 / 5
S=1/2acsinB=12/5