In the triangle ABC, given C = 10, B = 6, angle c = 120 degrees, then the value of sina is

In the triangle ABC, given C = 10, B = 6, angle c = 120 degrees, then the value of sina is

Cosine theorem A2 = B2 + C2 + 2bccosc = 36 + 100-60
a=√76=2√19
Zheng Xuan theorem
a/sinA=c/sinC
sinA=asinC/c=2√19*√3/(2*10)=√57/10

In △ ABC, if a = 4, B = 6, C = 120 °, then the value of edge C is ()
A. 8B. 217C. 62D. 219

In △ ABC, if ∵ a = 4, B = 6, C = 120 °, then C2 = A2 + b2-2ab · COSC = 16 + 36-48 × (- 12) = 76, ∵ C = 76 = 219 can be obtained from the cosine theorem

Triangle ABC, a = 5, B = 7, B = 120, find the value of sina and edge C

b/sinB=a/sinA
sinA=asinB/b=5sin120/7=5*√3/2/7=5√3/14
cosB=(a^2+c^2-b^2)/2ac=-1/2
a^2+c^2-b^2=-ac
25+c^2-49=-5c
c^2+5c-24=0
(c+8)(c-3)=0
c=3

It is known that in △ ABC, the opposite sides of the angle ABC are a, B and C respectively. If Sina + cosa = 2 and B = 2, B = π / 6, the value of a is obtained

Sina + cosa = √ 2, then √ 2Sin (a + 45 °) = √ 2, a = 45 °
Sine theorem a / Sina = B / SINB, a = (√ 2 / 0.5) × √ 2 / 2 = 2

In △ ABC, if a = 4, B = 6, C = 120 °, then the value of edge C is ()
A. 8B. 217C. 62D. 219

In △ ABC, if ∵ a = 4, B = 6, C = 120 °, then C2 = A2 + b2-2ab · COSC = 16 + 36-48 × (- 12) = 76, ∵ C = 76 = 219 can be obtained from the cosine theorem

In △ ABC, if a = 8, B = 4, C = 120 °, then the value of sina ∶ SINB is
Completion

Because C = 120
So a < 90, B < 90
According to the sine theorem a / Sina = B / SINB
So Sina / SINB = A / b = 2

In △ ABC, B = 120 °, a = 3, C = 5, then the value of sina + sinc is ()
A. 837B. 437C. 11233D. 5633

∵ in △ ABC, B = 120 °, a = 3, C = 5, then B2 = A2 + c2-2ac · CoSb = 9 + 25-30 × (- 12) = 49, ∵ B = 7 can be obtained from cosine theorem. Then 7sin120 ° = 3sina = 5sinc can be obtained from sine theorem, Sina = 3314, sinc = 5314, ∵ Sina + sinc = 437 can be obtained from simplification, so the answer is 437

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) (...)

The three inner angles a, B and C of △ ABC are opposite to a, B and C respectively. It is known that sina + sinc = 2sinb and B = π / 3. The area of △ ABC is (√ 3) / 2,
Then: B =? According to the specific degree of the answer, an additional reward of 5.50 points is offered. It's hard!

b=√ 2.
From the known Sina + sinc = 2sinb, a + C = 2B can be obtained by using sine theorem
If s, △ ABC = 1 / 2 * ac * SINB = (√ 3) / 2, AC = 2
From the complete square formula A & # 178; + C & # 178; = (a + C) &# 178; - 2Ac = 4B & # 178; - 4
Combined with cosine theorem CoSb = (A & # 178; + C & # 178; - B & # 178;) / (2Ac) = (3b & # 178; - 4) / 4 = 1 / 2, the solution is b = √ 2

It is known that in △ ABC, the diagonals of ABC on three sides are ABC, a + B = 2B, and Sina + sinc = 2sinb is proved

A / Sina = B / SINB = C / sinc = 2R, R is the radius of circumscribed circle of triangle ABC, which is proved by two equations