Answer:

Step-by-step explanation:
(This exercise is presented in Spanish and for that reason explanation will be held in such language)
El lado restante se determina por la Ley del Coseno:



Finalmente, el angulo C se halla por medio de la misma ley:




Answer:
this is not a real math equation.
Step-by-step explanation:
I think the answer is y=10
Answer:

Step-by-step explanation:
The given expression is :

It can be solved as follows :

So, the solution of the given expression is equal to
.