given sides,
a=36
b=45
c=27
From cosine law,
cosA=
[tex] \frac{ {b}^{2} + {c}^{2} - {a}^{2} }{2bc} \\ = \frac{ {45}^{2} + {27}^{2} - {36}^{2} }{2 \times 45 \times 27} \\ = \frac{3}{5} = 0.6[/tex]
or, A= cos^-1(0.6)=53.13°
similarly, cosB=
[tex] \frac{ {a}^{2} + {c}^{2} - {b}^{2} }{2ac} \\ = \frac{ {36}^{2} + {27}^{2} - {45}^{2} }{2 \times 36 \times 27} \\ = 0[/tex]
or, B= cos^-1(0)=90°
Since, B=90, the given triangle is right angled triangle.
