In △ ABC, if acosa = bcosb, judge the shape of △ ABC

∵ cosa = B2 + c2-a22bc, CoSb = A2 + c2-b22ac, ∵ B2 + c2-a22bc · a = A2 + c2-b22ac · B, which is reduced to: a2c2-a4 = b2c2-b4, that is, (A2-B2) C2 = (A2-B2) (A2 + B2), if A2-B2 = 0, a = B, then △ ABC is isosceles triangle; if A2-B2 ≠ 0, A2 + B2 = C2, this

Sine and cosine theorem in high school mathematics
If an internal angle of an obtuse triangle is π / 3 and the ratio of the maximum side length to the minimum side length is m, then the value range of M is B
A (1,2) B (2, positive infinity) C [3, positive infinity) d (3, positive infinity)
In △ ABC, cos & sup2; a / 2 = B + C / 2C, then the shape of triangle ABC is
In △ ABC, ∠ B = 30 ° AB = 2, root number 3, area s = root number 3, calculate AC
The answers to these questions should be explained in detail
I understand the third question. The second question is half a. the first question is not understood at all. Can you explain it in detail

1. Let B be 60 °, C be an obtuse angle, C be greater than or equal to 90 ° and a be less than or equal to 30 °, (the sum of internal angles is 180 ° and) angle c be an obtuse angle, so C is the largest, angle a is the smallest, and angle a is the smallest. (it can be seen from the large angle to the long side and small angle to the short side.) according to the sine theorem, C / sin C = A / sin a, C = 2 π / 3-A, so C / sin (2 π / 3-A) = a / sin a, C / a = si

In △ ABC, the edges of angles a, B and C are a, B, C, A2 + C2 − B2 = 12ac respectively. (I) find the value of sin2a + C2 + cos2b; (II) if B = 2, find the maximum area of △ ABC

(I) from the cosine theorem: CoSb = 14sin2a + C2 + cos2b = sin2 (π 2 − B2) + 2cos2b − 1 = cos2b2 + 2cos2b − 1 = 1 + cosb2 + 2cos2b − 1 = − 14 (II) from CoSb = 14, SINB = 154. ∵ B = 2, A2 + C2 − B2 = 12ac ∵ A2 + C2 = 12ac + B2 = 12ac + 4 ≥ 2Ac, so AC ≤ 83

In △ ABC, satisfy (2a-c) CoSb = bcosc
1. Find the size of angle B
2. Let m = (sin2a, cos2a), n = (4K, 1) (k > 1), and the maximum value of Mn be 5, then find the value of K

1、（2a-c）cosB=bcosC
Using the positive metaphysical theorem: 2sinacosb sinccosb = sinbcosc
Then, 2sinacosb = sinccosb + sinbcosc = sin (B + C) = Sina
A is not 0 or 180 degrees, so CoSb = 1 / 2; b = 60 degrees
2. Mn = 4ksin2a + cos2a = under the radical sign (16K * k + 1) sin (2a + P), tap = 1 / (4K)
So 16K * k + 1 = 25, so the solution!

In △ ABC, it is known that SINB + sinc = Sina (CoSb + COSC)______ ．

Let the opposite sides of a, B and C be a, B and C respectively. From SINB + sinc = Sina (CoSb + COSC), we can get: B + C = a (CoSb + COSC), CoSb = A2 + C2 − b22ac, COSC = A2 + B2 − c22ab, ∩ B + C = a (A2 + C2 − b22ac + A2 + B2 − c22ab), we can get: (B + C) (B2 + c2-a2) = 0, ∵ B + C ≠ 0

Sine and cosine theorems
The passenger ship sails at a constant speed from a to B and then to C at the speed of 2V. The cargo ship starts from the midpoint D of AC and sails at a constant speed along a straight line at the speed of V to deliver the cargo to the passenger ship. Ab ⊥ BC is known, and ab = BC = 50 nautical miles?
The figure is like this: B is the right angle of the triangle, AC is the hypotenuse, D is the midpoint of AC

Let the place where the two ships meet be x nautical miles from point C
By cosine theorem, de ^ 2 = CE ^ 2 + CD ^ 2-2 * ce * CD * cos 45 degree
=x^2+1250-50x.
AB+BE=50+(50-x)=100-x.
AB + be = 2DE,
4( x^2+1250-50x)=(100-x)^2,
x=50√6/3.
The two ships meet at a distance of 50 √ 6 / 3 nautical miles from point C

Let △ ABC be solved by the following conditions, where ()
A. b=20，A=45°，C=80°B. a=30，c=28，B=60°C. a=12，c=15，A=120°D. a=14，c=16，A=45°

A. According to the sine theorem Asina = bsinb, B = csinc, a = bsinasin, B = 102sin55 °, C = 20sin80 ° sin55 °, then the triangle has only one solution, this option is wrong; B, from a = 30, C = 28, B = 60 °, according to the cosine theorem, B2 = A2 + c2-2accosb = 844, the solution is b = 2211, that is, the triangle has only one solution, this option is wrong; C, from a = 12, C = 15, get a &If a = 14, C = 16, a = 45 ° according to sine theorem Asina = csinc, sinc = 16 × 2214 = 427 & gt; 22, C & gt; a, C & gt; 45 ° according to the image and property of sine function, C has two solutions, so D is selected

In △ ABC, a triangle is solved according to the following conditions, one of which is ()
A. b=7，c=3，C=30°B. b=5，c=42，B=45°C. a=6，b=63，B=60°D. a=20，b=30，A=30°

For a, SINB = bsincc = 76 ＞ 1 can be obtained from the sine theorem, at this time, the triangle has no solution, which is not suitable for the problem; for B, sinc = csinbb = 45, ∵ C ＞ B, B = 45 ° can be obtained from the sine theorem, at this time, C has two solutions, which is not suitable for the problem; for C, Sina = asinbb = 12, ∵ B ＞ a, B = 60 ° can be obtained from the sine theorem, at this time, a = 30 ° is suitable for the problem; for D, SINB = bsinaa = 3 can be obtained from the sine theorem 4, ∵ B ∵ a, a = 30 °, B has two solutions, which does not conform to the meaning of the question, so C

In △ ABC, if b2sin2c + c2sin2b = 2bccosbcosc, then △ ABC is ()
A. Equilateral triangle B. isosceles triangle C. right triangle D. isosceles right triangle

According to the sine theorem Asina = bsinb = csinc = 2R, a = 2rsina, B = 2rsinb, C = 2rsinc are obtained. Substituting into the known equation, (2rsinb) 2sin2c + (2rsinc) 2sin2b = 8r2sinbsinccosbcosc, that is, sin2bsin2c + sin2csin2b = 2sinbsincscosbcosc, sinbsincc ≠ 0

In the triangle ABC, a, B and C are the opposite sides of angles a, B and C respectively, a ^ 2-C ^ 2 = B ^ 2-8bc / 5, a = 3, and the area of ABC is 6
1. Find the sine value of angle a 2. Find the edge B and C

According to the cosine theorem, a ^ 2 = B ^ 2 + C ^ 2-2bc * cosa, then according to the condition, 2BC * cosa = 8bc / 5, then cosa = 4 / 5, then Sina = 3 / 5
According to the sine theorem, the area of ABC is 1 / 2 * Sina * BC = 6, then BC = 20, and a is known. Bring in the original equation, two unknowns of the two equations, and solve B = 4, C = 5 or B = 5, C = 4

In △ ABC, if acosa = bcosb, judge the shape of △ ABC