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."

enter image description here

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 ?

  • 1
    $\begingroup$ No, minimum angle between $BD,FD$ what is cos $BDF$? $\endgroup$ – Narasimham Mar 5 '17 at 20:33
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    $\begingroup$ Look at the triangle $BDF$ ... do a bit of Pythagorus to get the distances & then do some trig to get the angle at $B$. $\endgroup$ – Donald Splutterwit Mar 5 '17 at 20:34
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    $\begingroup$ Picture the triangle DBF, which is a right angled-triangle. The angle being referred to is the angle at D in this triangle $\endgroup$ – joesingo Mar 5 '17 at 20:34
  • $\begingroup$ Thanks for your help guys I have solved the question now :) $\endgroup$ – Dan Mar 5 '17 at 20:49

I see you have solved this in comments since I started typing, but I'll leave this here as an answer anyway)...

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.


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