Discharge at an angle in a channel concept question

Now I agree that the velocity of the discharge and the area is not at right angles, but I don't understand how using $$\cos \theta$$ helps? Shouldn't I use the area perpendicular to the channel? Like so:

We know that discharge $$Q$$ (also called flow rate) is given as the velocity of the fluid $$V$$ times the cross-section area $$A$$, which if written as an equation becomes $$Q = V * A$$
In the task, we are given a flow velocity $$V = 12 m/s$$ (inline with the pipe) and an area $$A = 60cm$$ measured along a vertical line. When given this information, we should immediately notice that the area $$A$$ is not perpendicular to the flow $$V$$, meaning that we cannot simply put $$Q = V * A = 7.2m^2/s$$. We need to decompose either the area or the velocity.
Procedure 1 Like you mentioned, we could use the cross-sectional area perpendicular to the pipe direction, which becomes $$A_1 = A * \cos \alpha$$. Why $$\cos \alpha$$? Well, this is due to the angle $$\alpha$$ as seen in the image below. The two angles are equal, but rotated 90 deg to each other. A method to spot such angle similarities is when the left "foot" of each angle is perpendicular to each other (when extending the lines) while at the same time the right "foot" of each angle is perpendicular to each other (when extening the lines). Having obtained the area $$A_1$$, we can combine it with the inline flow velocity $$V$$ giving $$Q = V * A_1 = V * (A * \cos \alpha) = 6.24 m^2/s$$
Procedure 2 (what the task did) Rather than thinking of the discharge inline with the pipe, consider the discharge through a vertical cross-section of the pipe. We may then use the area $$A$$, but must decompose the velocity to be perpendicular to $$A = 60cm$$. That is, we are interested in the x-component of the inline flow velocity (given a coordinate system x-y with x horizontal to the right). This gives $$V_x = V * \cos \alpha$$ as seen in the image below. Ultimately, we get $$Q = V_x * A = (V * \cos \alpha) * A = 6.24 m^2/s$$