Answer:[tex]98.13\ \text{or}\ 261.87^{\circ}[/tex]
Step-by-step explanation:
Given
[tex]\cos \beta =-\dfrac{\sqrt{2}}{10}[/tex]
As the value of cosine is negative, therefore the angle must lie in II or III quadrant
[tex]\Rightarrow \beta =\cos^{-1}(-\dfrac{\sqrt{2}}{10})\\\Rightarrow \beta=\cos^{-1}(-0.14142)\\\Rightarrow \beta =98.13^{\circ}\ \text{or}\ 270^{\circ}-8.13\\\Rightarrow \beta=98.13^{\circ}\ \text{or}\ 261.87^{\circ}[/tex]
