# What is the magnitude of the vector given below conditions?

Let vector B = 3m at 60 degree. Let vector C have the same magnitude as vector A. Vector C has a direction angle greater than of vector A by 25 degree. Dot product of A and B is 30. Dot product of B and C is 35. What is the magnitude and angle of vector A.

I know how to get the x and y components of vector B. I also know how the dot product is working but I'm stuck in forming equation so that I could get all their components. Any hint on how to attack this problem?

Let's assume that vectors A and C have magnitude A, and the angle of vector A is $\alpha$. Then, the angle of vector C is going to be $\alpha+25$. Here are all components: $$A_x=A\cos(\alpha)\\ A_y=A\sin(\alpha)\\ B_x=3\cos(60)\\ B_y=3\sin(60)\\ C_x=A\cos(\alpha+25)\\ C_y=A\sin(\alpha+25)$$ The dot product between A and B can be written as $$A\cdot B=A_xB_x+A_yB_y=3A(\cos(\alpha)\cos(60)+\sin(\alpha)\sin(60))=30$$ or $$A\cos(\alpha-60)=10$$ Similarly, for B and C: $$B\cdot C=B_xC_x+B_yC_y=3A(\cos(60)\cos(\alpha+25)+\sin(60)\sin(\alpha+25))=35$$ or $$3A\cos(35-\alpha)=35$$ From these equations, take the ratio, and you will get an equation in only $\alpha$, that you might need to solve numerically. Then just replace it into one of them, and you get A