Given the sine value of two angles, how to find the third angle

Given the sine value of two angles, how to find the third angle

From the sine theorem a = 2R * Sina B = 2R * SINB C = 2R * sinc, substitute the following cosine theorem C ^ 2 = a ^ 2 + B ^ 2-2ab * cosc4 (sinc) ^ 2 = 4 (Sina) ^ 2 + 4 (SINB) ^ 2 - (COSC) (Sina) (SINB) / 2 you say that given Sina and SINB, we can find out whether we can find C, it is better to replace it with COSC and find out the value of COSC, because C

Given the cosine value of two corners of a triangle, find the sine value of the other corner

Let the three angles of a triangle be a, B, C
Given cosa, CoSb, find sinc
SinC=Sin（180-A-B）=Sin（A+B）= SinA CosB+CosA SinB
Using cosa, CoSb to find the necessary conditions, we can find sinc

Finding the area of triangle by knowing the sine value of two corners and the length of the third side

Given SINB, sinc, a, then CoSb, COSC can be obtained
Sina = sin (180 - (B + C)) = sin (B + C) = sinbcosc + sinccosb
a/sinA=c/sinC c=asinC/sinA
S=acsinB/2=a^2sinBsinC/2sinA

Given that the sine value of the base angle B and C of the isosceles triangle ABC is 4 / 5, four trigonometric ratios of the vertex angle a are obtained

Let the waist length be 5 and the height 4, so that the bottom length is 6
The height of the waist is 24 / 5
sinA=24/25
cosA=7/25
tanA=24/7
cotA=7/24

Given the triangle ABC, ab = AC, angle BAC = 36 degrees, BD is the bisector of angle ABC, AC intersects with D, AE, vertical BC, perpendicular foot is e, find the value of sine 18 degrees

Let DF be perpendicular to AB, AB to F, BG to CD perpendicular to g, let BC = 2, then ad = BD = BC = 2, AF = BF = 2cos36 °. Then DG = GC = 2cos36 ° - 1. For angle BAE, sin18 ° = 1 / (4cos36 °), for angle GBC, sin18 ° = (2cos36 ° - 1) / 2

The problem of finding polar coordinates by vector in sinusoidal AC circuit Computer coordinates used in sinusoidal AC circuit A = x-jy is expressed in polar coordinates, and the angle φ = arctan Y / X when the modulus is root sign For example (30-j40) = 50 ∠ - 53 ° If I only know how to multiply one J40 by another polar coordinate 4.4 ∠ 73 j40×4.4∠73°

j40＝40∠90°
j40×4.4∠73°＝（40∠90°）×（4.4∠73°）
＝（40×4.4）∠（90°+73°）＝176∠163°
[multiplication formula: ρ 1 ∠ a °× ρ 2 ∠ B ° = ρ 1 ρ 2 ∠ (a + b) °]

When can Tellegen's theorem be used to analyze sinusoidal steady state circuit by vector method

Tellegen's theorem has no special requirements and is applicable to linear, nonlinear, time-varying and non-time-varying networks
However, the general analysis of sinusoidal steady-state circuit will not deliberately use Tellegen's theorem. Generally speaking, it is also used for complex power, but it seems that the apparent power is not satisfied. This is just my understanding~

Vector proof of sine theorem

Just now I see this proof, I do a unit vector I, I ⊥ BC in the triangle ABC plane, because Ba + AC + CB = 0 holds, I * Ba + I * AC = 0

Using vector method to express sine quantity can get two elements of sine quantity, which are and what are respectively?

This is my previous answer,
Copy as follows:
The three elements of sine, frequency, amplitude, and initial phase
A vector whose module length (length) is equal to the amplitude of the sine, the angle between its initial position and the positive direction of the horizontal axis is the initial phase of the sinusoidal quantity, and the vector is rotated anticlockwise at the angular frequency of the sinusoidal quantity. In this case, the vector also has three elements, which can be used to represent the sine quantity
In general, for example, in AC linear circuit, the frequency of voltage and current is the same and known
That is, as long as the amplitude of the sine and the initial phase can be obtained
In this case, we can use a vector to make the modulus length (length) equal to the amplitude of the sine, and make the angle between its initial position and the positive direction of the horizontal axis as the initial phase of the sinusoid

Why is the sine product of 20 degrees 40 degrees and 80 degrees the root three of eight Simplification of trigonometric function evaluation

20, 40, 80 looks awkward, right? It doesn't matter, does it? So I'm taking him a sin 60
sin20sin40sin80*(sin60)
=-1 / 2 (cos60-cos20) sin80 * (root 3 / 2)
=-1 / 4 * sin80 + 1 / 2 * cos20 * sin80 * Radix 3 / 2
=-1 / 4 * sin80 + 1 / 2 * [1 / 2 * (sin60 + sin100)] * radical 3 / 2
=(- 1 / 4 * sin80 + 1 / 4 * sin60 + 1 / 4 * sin100) * radical 3 / 2
=1/4*(sin60)^2
=1 / 4 * root 3 / 2 * root 3 / 2
=3/16
Then, calculate (3 / 16) / sin 60, which is the root of eight
This is the formula of product sum difference, which is mainly used to calculate the problem
sinαsinβ=-1/2[cos(α+β)-cos(α-β)]
cosαcosβ=1/2[cos(α+β)+cos(α-β)]
sinαcosβ=1/2[sin(α+β)+sin(α-β)]
cosαsinβ=1/2[sin(α+β)-sin(α-β)]