Alan

Assuming that by "legs" you mean thew sides that aren't the hypotenuse, then just use the arctan function.

That is, you have tan(A) = 50/45 so A = atan(50/45)  (i.e. tan^-1(A))

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{50}}}{{\mathtt{45}}}}\right)} = {\mathtt{48.012\: \!787\: \!504\: \!183^{\circ}}}$$

The other angle (B, say) is atan(45/50)

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{45}}}{{\mathtt{50}}}}\right)} = {\mathtt{41.987\: \!212\: \!495\: \!817^{\circ}}}$$

You should check that these two sum to 90°

c is twice as big as d.

2a + c = 180°

Each little dotted square in your graphs has sides of length 2 units.  This makes the radius of the circle in the top graph close to 5 units, so it's too big to be the one described by the equation.

What is the radius of the circle in the figure at the top?

In these cases the third components are zero (implicitly).  The vectors in full are:

p= -2i -3j+0k and q= 4i +7j+0k

Can't be the second one from the bottom as its centre has a positive x-value.

If you want to find the angle use atan(slope).  Unlike sin and cos, the tangent function is quite happy dealing with numbers greater than 1.

Well, from the equation we can see that the circle is centred at (-1.5, -1) and has a radius of 3.

Can you work out which one it is now?

Is O at the centre of the circle?  If so, then OA = OB, so OAB is an isosceles triangle, and the other two angles are the same.  If we call the other angles θ, then  2θ + 85 = 180, so 2θ = 95 and θ = 95/2 = 47.5°

Well, it's not a whole number because it has a fractional part.  Same for integer (another name for whole number)  It's not irrational because you can express it as the ratio of two whole numbers (25/9).

So it must be ...