Answer:
Since [tex]AB \neq BC \neq AC[/tex] and [tex]\cos B = 0[/tex], we conclude that given triangle is a right-angled and scalene.
Step-by-step explanation:
First, we determine the lengths of sides [tex]AB[/tex], [tex]BC[/tex] and [tex]AC[/tex] by the Equation of the Line Segment, which is based on the Pythagorean Theorem:
Line segment AB
[tex]AB = \sqrt{(4-1)^{2}+ (4-1)^{2}}[/tex]
[tex]AB = 3\sqrt{2}[/tex]
Line segment BC
[tex]BC = \sqrt{(6-4)^{2}+(2-4)^{2}}[/tex]
[tex]BC = 2\sqrt{2}[/tex]
Line segment AC
[tex]AC = \sqrt{(6-1)^{2}+(2-1)^{2}}[/tex]
[tex]AC = \sqrt{26}[/tex]
And the angles are found by means of the Law of Cosine:, where acute angles exist when [tex]0 < \cos \theta < 1[/tex], whereas obtuse angles exist for [tex]-1 < \cos \theta < 0[/tex].
Angle A
[tex]BC^{2} = AB^{2} + AC^{2} - 2\cdot AB\cdot AC \cdot \cos A[/tex] (1)
[tex]\cos A = \frac{BC^{2}-AB^{2}-AC^{2}}{-2\cdot AB\cdot AC}[/tex]
[tex]\cos A = \frac{8-18-26}{-2\dot (3\sqrt{2})\cdot (\sqrt{26})}[/tex]
[tex]\cos A = 0.832[/tex]
Angle B
[tex]AC^{2} = AB^{2} + BC^{2} - 2\cdot AB\cdot BC\cdot \cos B[/tex] (2)
[tex]\cos B = \frac{AC^{2}-AB^{2}-BC^{2}}{-2\cdot AB\cdot BC}[/tex]
[tex]\cos B = \frac{26-18-8}{-2\cdot (3\sqrt{2})\cdot (2\sqrt{2})}[/tex]
[tex]\cos B = 0[/tex]
Angle C
[tex]AB^{2} = AC^{2}+BC^{2}-2\cdot AC\cdot BC\cdot \cos C[/tex] (3)
[tex]\cos C = \frac{AB^{2}-AC^{2}-BC^{2}}{-2\cdot AC\cdot BC}[/tex]
[tex]\cos C = \frac{18 -26-8}{-2\cdot (\sqrt{26})\cdot (2\sqrt{2})}[/tex]
[tex]\cos C = 0.555[/tex]
Since [tex]AB \neq BC \neq AC[/tex] and [tex]\cos B = 0[/tex], we conclude that given triangle is a right-angled and scalene.
