If a × 0.8 = B △ 1.8 = C × 1 (a, B, C are all greater than 0), then the order of a, B, C is______ ＞______ ＞______ ．

If a × 0.8 = B △ 1.8 = C × 1 (a, B, C are all greater than 0), then the order of a, B, C is______ ＞______ ＞______ ．

A × 0.8 = 1A = 1.25b △ 1.8 = 1b = 1.8c × 1 = 1C = 1, so b > a > C

In the triangle ABC, the corresponding sides of three internal angles a, B and C are a, B and C respectively. Given that C = 3, C = 60 degrees and a + B = 5, then the value of COS ((a-b) / 2) is?
Hurry, hurry, hurry, hurry

According to the sine theorem and the equal ratio theorem
a/sinA=b/sin B=c/sinC=(a+b)/( sinA+ sin B)
Substituting C = 3, C = 60 ° and a + B = 5, 3 / sin 60 ° is 5 / (sin a + sin b)
sinA+ sin B=5√3/6.
2sin((A+B)/2)cos((A-B)/2) =5√3/6.
2×sin60°cos((A-B)/2) =5√3/6.
∴cos((A-B)/2) =5/6.

In the triangle ABC, a = 45 degrees, cos = 4 / 5. Find the value of sinc. If BC = 10, find the triangle area

Sinc = sin (a + b) = sinacosb + cosasinb = 7 / 10 times root sign 2
According to the sine theorem, we know AB = BC * sinc / Sina = 14
S=1/2AB*BC*sinB=42

In the triangle ABC, a = 2, B = 5, C = 6, then what is cos equal to?

cosa=(57/60) cosb=(5/8) cosc=(-7/20)

In the triangle ABC, if π / 3 ≤∠ B ≤ π / 2, then why is cos [(A-C) / 2] less than or equal to 1
The original problem is as follows - [in the triangle ABC, we know that π / 3 ≤∠ B ≤ π / 2, and prove a + C ≤ 2B]
I searched a lot of answers, but I didn't understand one place
They all came up with a formula: "cos [(A-C) / 2]

The range of cosine function f (x) = cosx is [- 1,1], so no matter what x is, the value of cosx cannot be greater than 1

In △ ABC, b2-bc-2c2 = 0, a = 6, cosa = 78, then the area of △ ABC is ()
A. 15B. 152C. 2D. 72

From the factorization of b2-bc-2c2 = 0, we get: (b-2c) (B + C) = 0, the solution is: B = 2c, B = - C (rounding off). According to the cosine theorem, we get: cosa = B2 + C & nbsp; 2 & nbsp; − A & nbsp; 22bc = B2 + C & nbsp; 2 & nbsp; − 62bc = 78, we get: 4B2 + 4c2-24 = 7bc, we substitute C = B2 into: 4B2 + b2-24 = 72b2, that is, B2 = 16, we get: B = 4 or B = - 4 (rounding off), then B = 4, so C = 2; We can get Sina = 158, so the area of △ ABC is 12bc · Sina = 152, so we choose B

The perimeter of triangle ABC is 20, the area is 10 triangles, and the angle a is 60 degrees?

S = 1 / 2R (a + B + C) = 1 / 2bcsin60 ° = 10 √ 3 (where R is the radius of the inscribed circle of the triangle) a + B + C = 20, so r = √ 3, let the center of the triangle be o, connect Ao, then bisector angle a, that is, angle oad = angle OAE = 30, make OD ⊥ AB, OE ⊥ AC, then ad = AE = 3 (right triangle, R and the corresponding angle are known) let BD =

In the triangle ABC, the known area is 16 3 / 3, a = 6, a, a = 60 degrees, then the perimeter of the triangle ABC is

Let AB = C, AC = B. first of all, you can get the first equation by using cosine theorem: cos60 ° = (b ^ 2 + C ^ 2-A ^ 2) / 2BC. Then, in the triangle ABC, make AC vertical line through B to intersect AC at point E. in the triangle Abe, you can find be = C times root 3 / 2, so according to the area in the topic, we know: 16 times root 3 / 3 = BC times root 3 / 4, so BC = 64 /

In the triangle ABC, the known area is 3 / 16 3, a = 6, a, a = 60 degrees, then the perimeter of the triangle ABC is

First, we use the area formula s = 1 / 2absin60 to get the length of B
Then we use cosine theorem cos60 = (b ^ 2 + C ^ 2-A ^ 2) / 2BC to find the length of C
Finally, find out the perimeter
(your area is 3 / 16 √ 3 or 3 √ 3 / 16, so it doesn't count. The process is given and you can apply it.)

As shown in the figure, in the triangle ABC, ab = AC, the center line on AC divides the perimeter of the triangle into two parts: 24cm and 30cm, and calculates the length of each side of the triangle ABC

AB = AC = 16 BC = 22 or AB = AC = 20 BC = 14