It is known that in △ ABC, Sina cosa = 7 / 5 (1) Sina × cosa (2) to judge whether △ ABC is an acute angle or an obtuse angle triangle

It is known that in △ ABC, Sina cosa = 7 / 5 (1) Sina × cosa (2) to judge whether △ ABC is an acute angle or an obtuse angle triangle

The two sides are square
sinAcosA=-12/25
Obtuse triangle,
（tanA)/(1+tan²A)=-12/25
The solution is Tana = - 3 / 4, or Tana = - 4 / 3

In △ ABC, if Sina + cosa = 1 / 5, then the shape of △ ABC is

It is known that sina + cosa = 1 / 5, (Sina) ^ 2 + (COSA) ^ 2 = 1, we can get: Sina · cosa = (1 / 2) {(Sina + COSA) ^ 2 - [(Sina) ^ 2 + (COSA) ^ 2]} = - 12 / 25 < 0, because 0 < a < π, we can get: Sina > 0, so, cosa < 0, we can get: π / 2 < a < π, that is: △ ABC

In the triangle ABC, a, B and C are the opposite sides of the angles a, B and C respectively, and Sina + cosa = C / B, find the angle B

sinA＋cosA=c/b=sinC/sinB
sinAsinB＋cosAsinB=sinC=sin(A＋B)=sinAcosB＋cosAsinB
sinAsinB=sinAcosB
sinB=cosB
tanB=1
B=45°

What triangle is Sina * cosa = 0 in triangle ABC

If a is an internal angle, then Sina ≠ 0
So cosa = 0
So a = 90 degrees
So it's a right triangle