First, let's make a diagram to visualize the problem.
According to the law of reflection, we know that
![\alpha=65](https://img.qammunity.org/2023/formulas/physics/college/8thc18yhc9s5jxrsrclbzjksefmaa91njt.png)
Also, these two angles are complementary so,
![\alpha+\lambda=90](https://img.qammunity.org/2023/formulas/physics/college/s3a4rjhmswctj2v87avdavjh5w1hd2wxl0.png)
Let's find lambda.
![\begin{gathered} 65+\lambda=90 \\ \lambda=90-65 \\ \lambda=25 \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/9uw71m6je82c3tfi30fwgpy6v0lzkirc0n.png)
Now, we can find angle gamma using the interior angles of a theorem (triangle).
![\begin{gathered} \lambda+120+\gamma=180 \\ 25+120+\gamma=180 \\ \gamma=180-145 \\ \gamma=35 \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/j4vs0vjbe4sg5suowb4e76ks0fq0i15tv4.png)
Then, we observe that angle gamma and angle beta are complementary, so let's find beta.
![\begin{gathered} \gamma+\beta=90 \\ 35+\beta=90 \\ \beta=90-35 \\ \beta=55 \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/dqbnqx8agta09enrpvlsu1jyhg9afy42ay.png)
At last, by the law of reflection, we state that
![\theta=\beta=55](https://img.qammunity.org/2023/formulas/physics/college/9g6mmja50csanuzcrqqn2mqpmkctewpwrm.png)
Therefore, the direction of the ray after its reflection on mirror M2 is 55.