# Using 3D trigonometry to find an angle

I have the question "Calculate the angle between the line DF and the plane ABCD in the cuboid pictured below, giving your answer to 1 decimal place."

I know how to pretty much solve this. I just don't understand which angle the question is referring too.

Is it referring to the angle labelled B ?

• No, minimum angle between $BD,FD$ what is cos $BDF$? – Narasimham Mar 5 '17 at 20:33
• Look at the triangle $BDF$ ... do a bit of Pythagorus to get the distances & then do some trig to get the angle at $B$. – Donald Splutterwit Mar 5 '17 at 20:34
• Picture the triangle DBF, which is a right angled-triangle. The angle being referred to is the angle at D in this triangle – joesingo Mar 5 '17 at 20:34
• Thanks for your help guys I have solved the question now :) – Dan Mar 5 '17 at 20:49

The angle you need is $\angle BDF$. The reason that is the correct angle is because $B$ is the nearest point on the plane to $F$, since the normal to the plane at $B$ goes through $F$.
You can use the Pythagorean theorem to get the value of $DB$, then take the arctangent for your result.