In △ ABC, a, B and C are the opposite sides of angles a, B and C respectively, a = 60 ° and (b-2c) / ACOS (60 ° + C)=

In △ ABC, a, B and C are the opposite sides of angles a, B and C respectively, a = 60 ° and (b-2c) / ACOS (60 ° + C)=

If CD is perpendicular to AB and D, then b-2c = - 2 (C-B / 2) = - 2dB
Then ACOS (60 ° + C) = ACOS (180 ° - b) = - acosb = - dB
So (b-2c) / ACOS (60 ° + C) = - 2dB / - DB = 2

It is known that in RT △ ABC, the angle c = 90 ° and P is any point on ab. PE ‖ AC intersects BC at point E and PF ‖ BC intersects AC at point F
The point m and N are the midpoint of PA and Pb respectively
Try to be as detailed as possible, using the eighth grade solution

Prove: in right triangle AFP, crossing point m is parallel line mg of PF, angle GMF = angle MFP. So triangle FGM is similar to triangle AFP. So angle MPF = angle GMF = angle MFP. So MF = MP. Similarly, NE = NP. So Mn = MP + NP = MF + ne

In RT △ ABC, ∠ BAC = 90 °, ab = 3, AC = 4, P is a moving point on edge BC, PE ⊥ AB is in E, PF ⊥ AC is in F, M is the midpoint of EF, then the minimum value of AM is___ ．

If AP is the height of the hypotenuse of the right triangle ABC, i.e. AP ⊥ BC, then am has the minimum value, and am = 12ap. According to Pythagorean theorem, BC = AB2 + ac2 = 5, ∵ s △ ABC = 12ab · AC = 12bc · AP

In known triangle ABC, ab = AC, ad is perpendicular to plane ABC, and CE = 2ad, prove plane BDE and vertical plane BCE

Connect ed, extend ed, Ca at point F, connect bf because ad vertical plane ABC, EC vertical plane ABC, so ad / / EC because CE = 2ad, so ad is the median line of triangle FCE, so AF = AC because AB = AC, so AB = AF = AC, so the angle FBC = 90 degrees because EC vertical plane ABC, FB in plane ABC, so EC vertical

In the triangle ABC, Tan = - 5 / 12 cos =?
Does it depend on the quadrant? But there are two cases when Tan is negative, in the second or fourth quadrant

No, the internal angle belongs to (0, π)
tanA

In △ ABC, the problem of proving: (a + b) ^ 2 * sin ^ 2 (C / 2) + (a-b) ^ 2 * cos ^ 2 (C / 2) = C ^ 2 has confused me and made the steps clear

Left = (a + b) & 178; sin & 178; (C / 2) + (a-b) & 178; Cos & 178; (C / 2)
=a²（sin²C/2+cos²C/2)+b²(sin²C/2+cos²C/2)+2ab(sin²C/2-cos²C/2)
=a²+b²-2abcosC
=The last step is the cosine theorem
=Right

Proof: in any triangle ABC, there must be sin "(a + b) △ 2" = cos [C △ 2]

A+B+C=π
So a + B = π - C
(A+B)/2=π/2-C/2
sin[(A+B)/2]=sin(π/2-C/2)=cos(C/2)
Don't know how to send a message to ask me

If sin θ + cos θ = - 1 / 5 and θ belongs to (0, π), then the value of cot θ is?

∵sin²θ+cos²θ=1;
The result is: sin θ + cos θ = - 1 / 5; the square of both sides leads to 1 + 2sinasinb = 1 / 25
sinAsinB=-12/25
The solution is Sina = 3 / 5; SINB = - 4 / 5 or Sina = - 4 / 5; SINB = 3 / 5
And θ belongs to (0, π), х Sina > 0
That is, Sina = 3 / 5; CoSb = - 4 / 5
∴cotθ=-4/3

Given sin β + cos β = 1 / 5, β∈ (0, pie), find the value of COS (β - 60 °) + cot β
How do you do it?
Process!

First, we get cos β by the first formula
Root sign (square of 1-cos β) + cos β = 1 / 5
If the equation is easy to solve, I'm puzzled. Let's talk about how to determine cos β
Because sin β + cos β

Let sin @ + cos @ = 1 / 5, @ belong to (0, pie) to find the value of cot @
Trigonometric function problem

∵sinα+cosα=1/5 ①
∴(sinα+cosα)^2=1+2sinα*cosα=1/25
∴sinα*cosα=-12/25