How to get ##\ddot r## when you have ##r##, ##\theta## and right trig

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Homework Statement
A radar detects that a particle, located ##40 m## above the ground, moves towards it with velocity ##v_r=5 \frac{m}{s}##. The modulus of ##\vec v## is constant, and the modulus of the acceleration at that moment is ##a=10 \frac{m}{s}##. The radius vector that goes from the radar to the particle forms an angle of ##60°## with the ground. Find ##\ddot r##
Relevant Equations
##sin(x)=\frac{op}{hyp}##
I have a right triangle: one of the angles is ##60°## (that's ##\theta##), one of the sides is ##40 m## long, and the hypotenuse is equal to the radius. Now I can find an expression for ##r## and that expression is ##r=\frac{height}{sin \theta}##. If I differentiate it, I'll get ##\dot r## and if I differentiated it again I would get ##\ddot r##. Now, how can I differentiate that expression? because time doesn't appear in the expressions.
 
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Like Tony Stark said:
Homework Statement: A radar detects that a particle, located ##40 m## above the ground, moves towards it with velocity ##v_r=5 \frac{m}{s}##. The modulus of ##\vec v## is constant, and the modulus of the acceleration at that moment is ##a=10 \frac{m}{s}##. The radius vector that goes from the radar to the particle forms an angle of ##60°## with the ground. Find ##\ddot r##
Homework Equations: ##sin(x)=\frac{op}{hyp}##

I have a right triangle: one of the angles is ##60°## (that's ##\theta##), one of the sides is ##40 m## long, and the hypotenuse is equal to the radius. Now I can find an expression for ##r## and that expression is ##r=\frac{height}{sin \theta}##. If I differentiate it, I'll get ##\dot r## and if I differentiated it again I would get ##\ddot r##. Now, how can I differentiate that expression? because time doesn't appear in the expressions.
This looks like a continuation of a problem you posted earlier. In that problem, I think you were looking for the angle between the velocity vector and a vector that is perpendicular to the hypotenuse of your right triangle. If you know that angle and the target's acceleration vector, then you can decompose the acceleration vector into two orthogonal components, one of which is the ##\ddot r## that you are looking for here. Edit: This assumes that the target's motion and the radar are all in the same plane.
 
tnich said:
This looks like a continuation of a problem you posted earlier. In that problem, I think you were looking for the angle between the velocity vector and a vector that is perpendicular to the hypotenuse of your right triangle. If you know that angle and the target's acceleration vector, then you can decompose the acceleration vector into two orthogonal components, one of which is the ##\ddot r## that you are looking for here. Edit: This assumes that the target's motion and the radar are all in the same plane.

Yess, it's another part of the same problem. Because I could do the first one (the one of the angle) with your help but I got stuck in the other one. The thing is that I don't know how to get the angle formed by the acceleration and the axis
 
Like Tony Stark said:
Yess, it's another part of the same problem. Because I could do the first one (the one of the angle) with your help but I got stuck in the other one. The thing is that I don't know how to get the angle formed by the acceleration and the axis
You know the magnitude of the velocity is constant. What does that tell you about the angle between the velocity vector and the acceleration vector?