In the triangle ABC, the opposite sides of ∠ a, B and C are a, B and C respectively, and C = 3 / 4 π, Sina = 5 / 5 root? If AB is equal to the root two of double, find the value of a and B?

In the triangle ABC, the opposite sides of ∠ a, B and C are a, B and C respectively, and C = 3 / 4 π, Sina = 5 / 5 root? If AB is equal to the root two of double, find the value of a and B?

(1) sinA=√5/5C=3π/4cosA=2√5/5A+B=π/4sinB=sin(π/4-A)=√2/2(cosA-sinA)=√10/10(2) a/sinA=b/sinB=c/sinCa=bsinA/sinB=(b×√5/5)/(√10/10)=b√2ab=b²√2=2√2b=√2a=2
The known set a = {x 2a-2}
A = {x | 2a-2 ﹤ x ﹤ a} B = {x | 1 ﹤ x ﹤ 2} CRB = {x | x ≤ 1 or X ≥ 2} because a is really included in CRB. The following classification is discussed: (1) if a is an empty set, it naturally conforms to so 2a-2 ≥ a, that is, a ≥ 2. (2) if a is not an empty set, then '2a-2 ﹤ a' and 'a ≤ 1 or 2a-2 ≥ 2' so a ≤ 1
If 2a-2 > = a, a is an empty set, then a > = 2, or 2a-2 > = 1 and a
In triangle ABC, angle a = 30 °, B = 12, triangle area is 18, then (Sina + SINB + sinc) divided by (a + B + C) is equal to what?
Suppose ABC circumcircle radius r,
There is a = 2rsina B = 2rsinb C = 2rsinc
c=2S/(b*sinA)=6
a^2=b^2+c^2-2bc*cosA=180-72√3
a=6√(5-2√3)
(sinA+sinB+sinC)/(a+b+c)
=(sinA+sinB+sinC)/[2R(sinA+sinB+sinC)]
=1/（2R）
2R*sinA=a
(sinA+sinB+sinC)/(a+b+c)
=1/2R
=sinA/a
=1/[12√(5-2√3) ]
1. Known set a = {x | X3
1. CRA intersection CRB = {x = 10} 2. A > = 7
In the triangle ABC, the angle a = 60 ° B = 1, C = 2, a + B + C divided by Sina + SINB + Inc
S = bcsina × 1 / 2 = C × radical 3 / 4 = radical 3
So C = 4
A = radical (B & sup2; + C & sup2; - 2bccosa) = 2 radical 3
a/sinA=b/sinB=c/sinC
2 radical 3 / (radical 3 / 2) = 1 / SINB = 4 / sinc
sinB=1/4,sinC=1
Original formula = (2 radical 3 + 4 + 1) / (radical 3 / 2 + 1 / 4 + 1) = 4
The known set a = x
A={x|x＜5},B={x|x＜a}
If a contains B
Then a ≤ 5
CrA={x|x≥5},CrB={x|x≥a}
If CRA does not contain CRB
Then a < 5
If you don't understand, please hi me, I wish you a happy study!
It is known that a, B and C are the three internal angles of △ ABC and satisfy 2sinb = Sina + sinc. Let B have the maximum value of B0. (I) find the size of B0; (II) find the value of Cosa COSC when B = 3b04
According to the cosine theorem, CoSb = A2 + C2 − b22ac = A2 + C2 − (a + C2) 22ac = 3 (A2 + C2) − 2ac2ac ≥ 6ac − 2ac8ac = 12. (4 points) because y = cosx monotonically decreases on (0, π), the maximum value of B is B0 = π 3. (6 points) (...)
Given a = {x ﹤ a}, B = {1 ﹤ x ﹤ 2}, and Au (CRB) = R, then the value range of real number a is?
There's a mistake upstairs
∵B={x|1
a≤2
CuB= ﹛x≤1,x≥2﹜
From Au (CRB) = R
a≥2
Not careful, alas
Because CRB is greater than or equal to 2 or less than or equal to 1, because the intersection of set a and CRB is r, so a should be greater than or equal to 2. You can draw the number axis on it
A. A ≤ 1 B.A ＜ 1 C.A ≥ 2 D.A ＞ 2 u is the Union (CRB) is the complement of B D D a > = 2
In the triangle ABC, the difference between the maximum angle c and the minimum angle a is 90 degrees, and Sina + sinc = 2sinb is used to find a: B: C
C-A = 90C = 90 + asinc = sin (90 + a) = cosa, so Sina + cosa = 2sinb square (Sina) ^ 2 + 2sinacosa + (COSA) ^ 2 = 4 (SINB) ^ 21 + sin2a = 4 (SINB) ^ 2 = 4-4 (CoSb) ^ 2C = 90 + AA + B + C = 2A + 90 + B = 1802a = 90-bsin2a = sin (90-b) = CoSb, so 1 + CoSb = 4-4 (CoSb) ^ 24 (co
Given the set a = {x | 2a-2 ＜ x ＜ a}, B = {x | 1 ＜ x ＜ 2}, and a ⊊∁ Rb, find the value range of A
∵ a = {x | 2a-2 ＜ x ＜ a}, B = {x | 1 ＜ x ＜ 2}, and a ⊊ Rb, ∁ RB = {x | x ≤ 1 or X ≥ 2}, ∁ a ≤ 1 or 2a-2 ≥ 2 or 2a-2 ≥ a, the solution is a ≤ 1 or a ≥ 2, then the range of a is {a | a ≤ 1 or a ≥ 2}