Question

Is the triangle obtuse, acute, equilateral or right?

Answer 1

Obtuse Bc Ik it is so ye lol

Answer 2

9514 1404 393

Answer:

  obtuse

Step-by-step explanation:

The law of cosines tells you ...

  b² = a² +c² -2ac·cos(B)

Substituting for a²+c² using the given equation, we have ...

  b² = b²·cos(B)² -2ac·cos(B)

We can subtract b² to get a quadratic in standard form for cos(B).

  b²·cos(B)² -2ac·cos(B) -b² = 0

Solving this using the quadratic formula gives ...

  $\cos(B)=\dfrac{-(-2ac)\pm\sqrt{(-2ac)^2-4(b^2)(-b^2)}}{2b^2}\\\\\cos(B)=\dfrac{ac}{b^2}\pm\sqrt{\left(\dfrac{ac}{b^2}\right)^2+1}$

The fraction ac/b² is always positive, so the term on the right (the square root) is always greater than 1. The value of cos(B) cannot be greater than 1, so the only viable value for cos(B) is ...

  $\cos(B)=\dfrac{ac}{b^2}-\sqrt{\left(\dfrac{ac}{b^2}\right)^2+1}$

The value of the radical is necessarily greater than ac/b², so cos(B) is necessarily negative. When cos(B) < 0, B > 90°. The triangle is obtuse.