The opposite sides of the inner angles a, B, C of the triangle ABC are a, B and C respectively, and cosa / 2 = two fifths of the root sign 5, AB vector point times AC vector is equal to 3 If B + C = 6, find a

The opposite sides of the inner angles a, B, C of the triangle ABC are a, B and C respectively, and cosa / 2 = two fifths of the root sign 5, AB vector point times AC vector is equal to 3 If B + C = 6, find a

First question:
∵cos（A/2）＝2√5/5,∴［cos（A/2）］^2＝4/5,∴cosA＝2［cos（A/2）］^2－1＝3/5＞0,
A is an acute angle, ν Sina = √ [1 - (COSA) ^ 2] = √ (1-9 / 25) = 4 / 5
∵ cosa = vector ab · vector AC / (︱ vector ab ︱ vector AC ︱ cosa = 3 / 5,
/ / vector ab · vector AC / (︱ vector ab ︱ vector AC) = 3 / 5,
∴3/（cb）＝3/5,∴bc＝5.
The area of △ ABC = (1 / 2) bcsina = (1 / 2) × 5 × (4 / 5) = 2
Second question: you may have made a mistake while you are busy! It should be to ask a
According to the cosine theorem, a ^ 2 = B ^ 2 + C ^ 2-2bccosa = (a + b) ^ 2-2bc-2bccosa,
∴a^2＝36－2×5－2×5×（3/5）＝26－6＝20,∴a＝2√5.
Note: if the second question is not my guess, please add

In △ ABC, the opposite sides of ∠ a, ∠ B and ∠ C are a, B, C respectively, and satisfy that a = 2 √ 5 of Cos2 and AC = 3 of ab × vector Find the area of (1) △ ABC (2) if B + C = 6, find the value of A Moreover, a = 2 √ 5 / 5, ab × vector AC = 3

A = 2 √ 5 / 5
Then cosa = 2cos? A-1 = 3 / 5
That is, Sina = 4 / 5
From ab × vector AC = 3
That is / AB / * / AC / * cosa = 3
That is / AB / * / AC / = BC = 3 / cosa = 5
The area of △ ABC = 1 / 2 * / AB / * / AC / * Sina = 1 / 2 * 5 * 4 / 5 = 2
2 a²=b²+c²-2bccosA
=（b+c）²-2bc-2bccosA
=（6）²-2*5-2*5*3/5
=20
That is, a = 2 √ 5

In △ ABC, we know that 2Ab * AC (AB, AC is a vector) = radical 3 * AB * AC, AB, AC is the module of vector) = 3bC ^ 2. Find the size of angle a, B, C

2Ab * AC (AB, AC is vector) = radical 3 * AB * AC AB, AC is the module of vector)
====>cosA=√3/2===>A=30º
Radical 3 * AB * AC AB, AC is the module of vector) = 3bC ^ 2
====>cb=√3a²===>sinCsinB=√3sin²A=√3/4
Sincsina = [cos (B-C)] / 2 - [cos (B + C)] / 2 = [cos (B-C)] / 2 + [cosa] / 2
=[cos(B-C)]/2+√3/4=√3/4
ν [cos (B-C)] / 2 = 0 = = > B-C = 90 ° or C-B = 90 °
B + C = 180-30 = 150 ° C
ν B = 120 ° C = 30 ° or B = 30 ° C = 120 ° C

In the triangle ABC, the angle a is 45 ° and ab = radical 6 BC = 2

BC/sinA=AB/sinC
sinC=√3/2
∠ C = 60 ° or 120 °
①∠C=60º ∠A=45°∠B=75°
Sine theorem AC = √ 3 + 1
②∠C=120°∠A=45°∠B=15°
AC=√3-1

In the known triangle ABC, a = root 3, B = root 2, B = 45 degrees?

A / Sina = B / SINB; √ 3 / Sina = √ 2 / (√ 2 / 2) = 2; Sina = √ 3 / 2; a = 60 or 120 degrees; a = 60, C = 75 degrees, C > A; a = 120, C = 15 degrees, C = 15 degrees

In the triangle ABC, we know that C = radical 2, B = 2 radical 3 / 3, B = 45 ° to solve the triangle

∵c/sinC=b/sinB
∴sinC=c*sinB/b=√3/2
∴C=π/3=60°
∴A=π-B-C=75°
∵a/sinA=b/sinB
∴a=b*sinA/sinB=√2
It's not easy to type,

In △ ABC, C = 2 times root sign 3 ∠ a = 45 degree a = 2 times root sign 6 solve triangle In △ ABC, ∠ a = 30 degrees ∠ B = 45 degrees, B = 12 solve triangles

The first problem: first of all, by using the sine theorem, a △ Sina = C △ sinc, we get that: sinc = half, so, ∠ C = 30 ° or 150 ° because a is a double root 6, which is larger than the length of C side. According to the principle of large side to large angle, ∠ C = 30 ° because ∠ a + ∠ C + ∠ B = 180 ° so ∠ B = 105 ° can be obtained

Given the triangle ABC, ab = 2 BC = root 10 AC = 3, find the value of vector AB * AC

Vector ab × vector AC = ab × AC × cos < vector AB, vector AC > 0
=2×3×（3^2+2^2-(√10)^2)/2×2×3
=3/2

In the triangle ABC, ab = 4, BC = 2, Radix 2, and Ba vector multiplied by BC vector = - 8, then AC = how much

CoSb = (vector Ba * vector BC) / (| ab | * | BC |) = - 8 / [4 * (2 √ 2)] = - 1 / √ 2. According to the cosine theorem, AC? = AB? + BC? - 2Ab * BC * CoSb = 4? + (2 √ 2)? - 2 * 4 * (2 √ 2) * (- 1 / √ 2) = 16 + 8 + 16 = 40 ﹤ AC = √ 40 = 2 √ 10

In △ ABC, ab = 3, AC = 2, BC= 10, then AB• AC=（　　） A. −3 Two B. −2 Three C. 2 Three D. 3 Two

∵ cosa = 9 + 4 − 10 is obtained from cosine theorem
2×3×2，
∴cos∠CAB＝1
4，
Qi
AB•
AC＝3×2×1
4＝3
2，
Therefore, D is selected