Given that angle a is the inner angle of triangle ABC, and SINB + C / 2 = root 3 / 2, find cosa

Given that angle a is the inner angle of triangle ABC, and SINB + C / 2 = root 3 / 2, find cosa

SINB + C / 2 = root 3 / 2 = sin60 degrees
So: (B + C) / 2 = 60 degrees
B + C = 120 degrees
Then: a = 60 degrees
So: cosa = 1 / 2

The opposite side of triangle A.B.C is a.b.c. it is known that cosa = 2 / 3. SINB = √ 5cosc. Find the value of Tanc. If a = √ 2. Find the ABC area of triangle

Find Tanc
∵cosA=2/3,∴sinA=√(1-cos ² A)=√5/3
∵sinB=√5cosC
sinB=sin(A+C)=sinAcosC+cosAsinC
∴sinAcosC+cosAsinC=√5cosC
∴√5/3cosC+2/3sinC=√5cosC
∴ sinC=√5cosC ,
∴tanC=√5
Find △ ABC area
When a = √ 2,
∵ sinA=√5/3
∴2R=a/sinA=√2/(√5/3)=3√10/5
∵ sinC=√5cosC,sin ² C+cos ² C=1
∴cos ² C=1/6,sin ² C=5/6,
sinC=√30/6,cosC=√6/6
∴sinB=√5cosC=√30/6
∴b=c=2RsinB=3√10/5*√30/6=√3
‡ area of triangle ABC
S=1/2*bcsinA=1/2*3*√5/3=√5/2

It is known that in triangle ABC, a, B and C are the opposite sides of angle ABC respectively, and satisfy cosa (root 3sina COSA) = 1 / 2 It is known that in triangle ABC, a, B and C are the opposite sides of angle ABC respectively, and satisfy cosa (√ 3sina COSA) = 1 / 2 1. Find the size of angle A 2. If a = 2 √ 2, s △ ABC = 2 √ 3, find the length of B.C Note '√' means root sign

1、
cosA·(√3sinA-cosA)=√3sinAcosA-cos ² A = √ 3 / 2sin2a - (1 + cos2a) / 2 = √ 3 / 2sin2a cos2a / 2-1 / 2 = sin (2A - π / 6) - 1 / 2 = 1 / 2, that is, sin (2A - π / 6) = 1, a = π / 3
2、
S = bcsina / 2 = √ 3 / 4bc = 2 √ 3, BC = 8
Cosine theorem cosa = (b) ²+ c ²- a ²)/ (2bc)=(b ²+ c ²- 8) / 16 = 1 / 2, B ²+ c ²= sixteen
∴b ²+ c ²+ 2bc=(b+c) ²= 32, B + C = 4 √ 2
b ²+ c ²- 2bc=(b-c) ²= 0, B = C
∴b=c=2√2

In triangle ABC, angle a = 60 degrees, B = 1, triangle area = root 3, find (a + B + C) / (Sina + SINB + sinc) I calculated C = 4

(a + B + C) / (Sina + SINB + sinc) = (2rsina + 2rsinb + 2rsinc) / (Sina + SINB + sinc) = 2R triangle area s = BC * Sina / 2 = root 3 * C / 4 = root 3, so C = 4A ^ 2 = B ^ 2 + C ^ 2-2bc * cosa = 13A = root 13, so a / Sina = root 13 / (root 3 / 2) = 2 root 39 / 3 = 2R, so (a + B + C) / (sin

In △ ABC, a = 60 °, B = 1, △ ABC area is 3, then a + B + C The value of sina + SINB + sinc is () A. 2 thirty-nine three B. 26 three three C. 8 three three D. 2 three

∵S△ABC=1
2bcsinA=1
two × one × c ×
three
2=
three
∴c=4
According to the cosine theorem: A2 = B2 + c2-2bccosa = 1 + 16-2 × one × four × one
2=13
So, a=
thirteen
According to sine theorem a
sinA=b
sinB=c
Sinc, then:
a+b+c
sinA+sinB+sinC=a
sinA=2
thirty-nine
three
So choose a

It is known that the perimeter of triangle ABC is root 2 plus 1, and Sina + SINB = twice the root. Sinc. (1) find the length of edge C (2 It is known that the perimeter of triangle ABC is root 2 plus 1, and Sina + SINB = 2 times the root sinc. (1) find the length of side C (2) if the area of triangle ABC is 1 / 6 times sinc, find the degree of angle C

(1) By the sine theorem a / Sina = B / SINB = C / sinc = 2R (R is the center of the circumscribed circle of triangle ABC) (AB = C, BC = a, AC = b)
Get: (A / 2R) + (B / 2R) = (√ 2) C / 2R
Because: a + B + C = 1 + √ 2
So: ab = C = 1
(2) Because: s triangle ABC = (1 / 2) absinc = (1 / 6) sinc
So: ab = 1 / 3
Because: a + B + C = 1 + √ 2, C = 1
So: a + B = √ 2
From the cosine theorem, COSC = (a ^ 2 + B ^ 2-C ^ 2) / (2Ab) = [(a + b) ^ 2-C ^ 2-2ab] / (2Ab) = [(a + b) ^ 2-C ^ 2] / (2Ab) - 1
=(2-1)/(2/3)-1=3/2-1=1/2
So: C = 60 °