In triangle ABC, sin is known ² A+sin ² B+sin ² C = 2, then the triangle is? emergency

In triangle ABC, sin is known ² A+sin ² B+sin ² C = 2, then the triangle is? emergency

sin ² A＋sin ² B＋sin ² C＝sin ² A＋sin ² B＋sin ² (A+B)＝sin ² A＋sin ² B＋(sinAcosB+cosAsinB) ²＝ sin ² A＋sin ² B＋sin ² Acos ² B＋cos ² Asin ² B＋2sinA...

In triangle ABC, ∠ a, ∠ B and ∠ C are respectively a, B and C, (1) it is proved by cosine theorem that when a ^ 2 + B ^ 2

1cosC=a^2+b^2-c^2/2ab
ab>0 cosC

The cosine theorem is used to explain that the necessary and sufficient conditions for the interior angle c of triangle ABC to be acute angle, right angle and obtuse angle are a^2+b^2>c^2、a^2+b^2=c^2、a^2+b^2

cos C = (a^2 + b^2 - c^2) / 2ab
Cos C0 for different angles

In the obtuse angle △ ABC, ∠ C is the obtuse angle. Try to deduce the sine theorem

Make the circumscribed circle O of △ ABC, make the diameter ad of O through point a, and connect BD. since the quadrilateral acbd is internally connected to the circle, ∠ ADB = flat angle - ∠ C, and obviously C = AB = ad sin ∠ ADB, C = 2R sin ∠ ADB = 2R sinc, i.e. C / sinc = 2R, where R represents the radius of O. the other two relations a / Sina = 2R and B / SINB = 2R

  If the triangle ABC is an obtuse triangle, how to prove the sine theorem?

As an auxiliary line, it is proved by the formula of sine

How to prove the sine theorem when the triangle ABC is an obtuse angle?

As shown in the figure, make a brief auxiliary line process. In the obtuse angle △ ABC, B is the obtuse angle, and the diameter of the circumscribed circle is recorded as 2R.