Triangle sine cosine theorem In the unequal edge △ ABC, a is the largest edge and a ^ 2

Triangle sine cosine theorem In the unequal edge △ ABC, a is the largest edge and a ^ 2

a^2＜b^2+c^2,
b^2+c^2-a^2＞0
So cosa = (b ^ 2 + C ^ 2-A ^ 2) / 2BC > 0
So a ＜ 90 °
And a is the largest side
So a > b, a > C
So 2A ＞ B + C = 180 ° - A
Therefore, 3A > 180 °
So a ＞ 60 °
So 60 ° < a < 90 °

Understand by sine theorem and cosine determination: in a known triangle ABC, edge a = 4, angle a = 45 degrees, angle B = 60 degrees, find edge B and the area of the triangle

Sine theorem: A / Sina = B / SINB
b=a*sinB/sinA=4*√3/2/(√2/2)=2√6
C=75°
sin(75°)=sin(30°+45°)=sin30°cos45°+cos30°sin45°=(√6+√2)/4
S=1/2*a*b*sinC
=6+2√3

Calculating the area with the known angle in the cosine theorem If the angle a is known to be 60 degrees in a triangle, the area s is? (represented by a, B, c)

In the sine theorem, there are
S△=(1/2)*ab*sinC,
And, (sinc) ^ 2 + (COSC) ^ 2 = 1, yes
sinC=√[1-(cosC)^2],
∴S△=(1/2)*ab*sinC=(1/2)*ab*√[1-(cosC)^2]=(√3/4)*ab.

In triangle ABC, ab = 6, angle a = 30 degrees, angle B = 120 degrees. Find the area by cosine theorem

∠C=180°-30°-120°=30°;
∴AB=BC=6;
Area = ab × BC × sin120° × (1/2)=6 × six × (√3/2) × (1/2)=9√3;
If you don't understand this question, you can ask,

Please prove that in triangle ABC: cos α= （sin ²γ+ sin ²β- sin ²α 　）/2sin γ* sin β

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In triangle ABC, sin ² A / 2 = (C-B) / 2C, judge the triangle shape

Because sin ² (A/2)=（c-b）/(2c)
So (1-cosa) / 2 = 1 / 2 - B / (2C)
That is, cosa = B / C
b=c*cosA
Then 2B ²= 2bccosA
Reason determined by cosine: a ²= b ²+ c ²- 2bccosA
That is, 2bccosa = B ²+ c ²- a ²
So 2B ²= b ²+ c ²- a ²
Then a ²+ b ²＝ c ²
The three sides of a triangle satisfy the Pythagorean theorem
So this triangle is a right triangle