Solving triangle problems in senior one In the triangle ABC, if (C / A + b) + (A / B + C) = B, then B =? There is a process to offer reward points

Solving triangle problems in senior one In the triangle ABC, if (C / A + b) + (A / B + C) = B, then B =? There is a process to offer reward points

The angle is B, so only using B & sup2; = A & sup2; + C & sup2; - 2accosb ① is enough. A & sup2; / (B + C) + C & sup2; / (a + b) = B is divided into a & sup3; + C & sup3; + B (A & sup2; + C & sup2;) = B (B & sup2; + AC + (a + C) B to meet the first B & sup2; a & sup3; + C

1. In the triangle ABC, if C + B = 6 (√ 3 + 1) a = 30 degrees, B = 30 degrees, find C =? B =?
2. Don't understand the following triangles
There are two solutions for a = 7, B = 14, a = 30 degree, B a = 30, B = 25, a = 150 degree
C a = 6 B = 9 A = 45 degree has two solutions D. B = 9 C = 10 b = 60 degree
3. In the triangle ABC, a = 60 degrees, BC = 3, then the circumference of the triangle ABC =?

1. According to the sine theorem (SINB) / b = (sinc) C, we can get C = 120 ° from the meaning of the problem, and the solution is C / 2 = (√ 3) B / 2, and C + B = 6 (√ 3 + 1), then B = 6, C = 6 (√ 3) 2. B. We have a corollary. Let's see if it's useful ---! When we know a, B and Sina, there are several cases: ① a > b, one solution: ② a = B, one solution: ③ 1 when a > sin

It is known that in △ ABC, a + C + 2B, AB + 1, BC + 4, find the length of the midline ad on the side of BC
It's from the chapter of sine and cosine theorem
It is known that in △ ABC, a + C = 2B, ab = 1, BC = 4. Find the length of the midline ad on the side of BC
== wrong number

∠A+∠B+∠C=360°
∠A+∠C=2∠B
∠B=60°
BD=BC÷2=2
AD^2=BA^2+BD^2-2×BA×BD×cos∠B
AD=√(5-2√3)
[√ denotes root]

The three sides of a triangle are the square of X + X + 1, the square of X - 1, 2x + 1 and the shape of a triangle

First of all, we need to judge the largest edge of the triangle; because (X & # 178; + X + 1) - (X & # 178; - 1) = x + 2 > 0, that is, X & # 178; + X + 1 > X & # 178; - 1, and (X & # 178; + X + 1) - (2x + 1) = x (x-1), known x > 1, that is, X (x-1) > 0, so x & # 178; + X + 1 > 2x + 1

In a triangle, we know that cos C / cos B = (3a-c) / B is sin B

cos C/cos B=（3a-c）/b
Using cosine theorem: [(a ^ 2 + B ^ 2-C ^ 2) / 2Ab] / [(a ^ 2 + C ^ 2-B ^ 2) / 2Ac] = (3a-c) / b
After simplification: 2Ac = 3A ^ 2 + 3C ^ 2-3b ^ 2
(a ^ 2 + C ^ 2-B ^ 2) / 2Ac = 1 / 3 (formula 1)
cosB=1/3
So SINB = (2 2) / 3

When the fishing boat was in distress, our navy immediately detected that the fishing boat was at C with an azimuth of 45 degrees and a distance of 10 nautical miles, and detected that the fishing boat was approaching Island b with an azimuth of 105 degrees and a speed of 9 nautical miles per hour. Our navy immediately went to catch up with the fishing boat with a speed of 21 nautical miles per hour to find the time required for the ship to find the fishing boat

From the cosine theorem: cos120 ° = (100 + 81t ^ - 441t ^) / (2 × 10 × 9t), t = 2 / 3 is obtained
According to the sine theorem, 9t / sin, θ = 21t / sin, 120 °, sin, θ = 3 √ 3 / 14, θ = acsin (3 √ 3 / 14), 45 ° - θ = 45 ° - acsin (3 √ 3 / 14)
The ship should move in the direction of 45 ° North acsin (3 √ 3 / 14) east, and the time to approach the fishing boat is 40 minutes

The solution of the exercises of sine and cosine theorem
In triangle ABC, angle a equals 45 degrees, ACA equals root sign 10, COSC equals root sign 5 of 2 / 5
(1) Find the length of BC
(2) Note that the midpoint of AB is D, and find the length of the centerline CD

sin C = √[ 1 - ( cos C )^2] = √5/5.
Sin B = sin (135 degrees - C) = 3 √ 10 / 10
According to the sine theorem, it can be concluded that:
BC / sin a = AC / sin B, so BC = AC sin a / sin B = 5 √ 2 / 3
Similarly, AB / sin C = AC / sin B, ab = 2 √ 5 / 3; thus, ad = √ 5 / 3
Using the cosine theorem, the center line CD can be obtained

Exercises of cosine theorem
In triangle ABC, if a ^ 4 + B ^ 4 + C ^ 4 = 2C ^ 2 (a ^ 2 + B ^ 2), then angle c is equal to
A.30
B.45
C. 45 or 135
D. 30 or 150

A problem of sine and cosine theorem
A plane flies from city a to city B at the speed of 326km / h along the course of 75 degrees north by East. 18 minutes later, due to weather reasons, the plane changes to another city C. what course should the plane follow when receiving the order? What is the distance from City C at this time?
I don't know where C is, and there is no picture in the title

Where is C?
"High school discussion ability is also very important
The score is high
You can discuss all the positions of C“
You don't have to talk nonsense. You can't form a definite triangle on one side. You have to discuss the infinite range. It's unnecessary

Application of sine and cosine theorem
It is known that in Δ ABC, a = 45 degrees, ab = radical 6, BC = 2

BC:sin45°=AB:sinC
Sinc = √ 3 / 2
Therefore, C = 60 ° or 120 °
If ∠ C = 60 °, then ∠ B = 180-60-45 = 75 °
AC = √ (6 + 2 * 2-2 * radical 6 * 2 * cos75 °)
=1+√3
If ∠ C = 120 °, then ∠ B = 180-60-45 = 15 °
AC = √ (6 + 2 * 2-2 * radical 6 * 2 * cos15 °)
=√3 -1