Calculating the opposite side of two triangles on the x and y axis with a correction for yaw

For a project I want to do a very simple Pythagoras calculation in C++. An object is equiped with an IMU sensor that gives either a Quaternion rotation or Euler angles. What I want to know is the opposite sides of the triangle underneath the object.:

I want to know these sides of the triangle for both the X and Y axis (black arrows):

This is pretty much very simple, except for the fact that the object can rotate. When the object is rotated I still want to use the X and Y axis in world space (black arrows), but when yawing the Euler angles of the IMU provide me with pitch and roll, which are in local space (red arrows):

In what way can I still get the world space angles (black arrows) while yawing, to be able to calculate my simple Pythagoras calculation? If I can't get them, is there a way to calculate the opposite sides I want using Quaternions?

Solution

We can do the calculation by taking into account the Euler angles in the following order -

Pitch

First of all, as you change the roll of the sensor, the sensor "ray" sweeps out a plane inclined to the horizon at angle pitch. We need to first calculate the closest distance between (i) the line of intersection between the plane and the ground, and (ii) the point directly below the sensor on the ground. This is given by d = h * tan(pitch).

Roll

Next we need to do another trigonometric step. As before the roll sweeps through the plane. The offset distance along the axis perpendicular to the line joining (i) and (ii) is given by f = h / cos(pitch) * tan(roll). This gives the intersection point on the ground to be (d, f)

Yaw

Previously, we considered a frame in which the yaw was zero. We now need to rotate this intersection point around the Z-axis by yaw. Thus the final intersection point is given by (x, y) = (d * cos(yaw) - f * sin(yaw), d * sin(yaw) + f * cos(yaw)). You can calculate the "space angle" you want by taking atan2(y, x).