Answer:
A.[tex]Cos\theta<0[/tex]
B.The triangle is not a right triangle
C.[tex]\theta[/tex] is an obtuse angle
Step-by-step explanation:
We are given that a triangle with sides a, b and c.
[tex]a^2+b^2<c^2[/tex]
When a triangle is an obtuse triangle then
[tex]a^2+b^2<c^2[/tex]
It means given triangle is an obtuse triangle .
Obtuse triangle is that triangle in which one angle is an obtuse angle.
Let [tex]\theta[/tex] be the measure of angle opposite the side of length c.
We have to find the statements which are must be true.
Cosine law:
[tex]c^2=a^2+b^2-2abCos\theta[/tex]
Substitute the value then we get
[tex]c^2<c^2-2abCos\theta[/tex]
Subtracting [tex]c^2[/tex] on both sides of inequality
[tex]c^2-c^2<c^2-c^2-2ab Cos\theta[/tex]
[tex]0<-2ab Cos\theta[/tex]
Adding [tex]2abCos\theta[/tex] on both sides then we get
[tex]2abCos\theta<-2abCos\theta+2abCos\theta=0[/tex]
[tex]2abCos\theta<0[/tex]
[tex]Cos\theta<0[/tex]
[tex]Cos\theta[/tex]is negative in second quadrant and third quadrant.
Therefore, [tex]\frac{\pi}{2}<\theta<\frac{3\pi}{2}[/tex]
But, we know that an obtuse angle is that angle which is greater than 90 degrees and less than 180 degrees.
Hence, [tex]\frac{\pi}{2}<\theta<\pi[/tex]
Option A ,C and D are true.
