Calculating Angle C in Vertical Plane: Trigonometric Problem in Dynamics Book

[Pyrrhus]

Hello, I'm in need of a hint or few pointers on how to calculate the angle C of the picture attached. I've already calculated y.

I was doing a few problems in this Dynamics book, i bought recently, and the ascention angle (angle C) is beating me

"The airplane C is being tracked down by the radar stations A and B. At the instant shown on the picture, the triangle ABC encounters itself in vertical plane and the lectures are Angle A = 30 degrees, Angle B = 22 degrees, Angular Speed A = 0.026 rad/s, Angular Speed B = 0.032 rad/s. Find a) the height y, b) the magnitude of the velocity (the vector V is at point C directed at an ascention angle (angle C) with respect tot he horizontal c) the ascention angle at the instant shown (angle c)"

ah yes distance d = 1000 m and it's between the stations A and B.

I hope the diagram is clear enough...

 

[arildno]

A few hints:
1) Having "y", it is easy determine lengths of AC, BC, and form the vector from A to C, and the vector from B to C.
Let for example the vector from AC have the form \vec{r}_{AC}=r_{AC}\hat{r}_{AC}
where r_{AC},\hat{r}_{AC} are the length and direction vector, respectively.

2) Let \hat{n}_{AC} be the unit vector in the plane of the triangle perpendicular to \hat{r}_{AC} and pointing in the direction of increasing angle, and make a similar construction for \hat{n}_{BC}

3) Decompose your velocity as:
\vec{v}=v_{AC}\hat{r}_{AC}+v_{BC}\hat{r}_{BC}

4) We therefore have, for example the equality:
\vec{v}\cdot\hat{n}_{AC}=r_{AC}\omega_{AC}\to{v}_{BC}\hat{r}_{BC}\cdot\hat{n}_{AC}}=r_{AC}\omega_{AC}\to{v}_{BC}=\frac{r_{AC}\omega_{AC}}{\hat{r}_{BC}\cdot\hat{n}_{AC}}
where \omega_{AC} is the angular velocity measured at A.

5) Thus, we have determined \vec{v} and may answer the two remaining questions.
Remember that \hat{r}_{AC},\hat{r}_{BC} are not orthogonal vectors!

 

[Pyrrhus]

Thanks Arildno, i was able to solve it.