Angle Between a Line and a Plane

Steps

1. 1

   Drop: from the top of the line, go straight down to the plane.

2. 2

   Join: from that foot to where the line meets the plane. This is the projection.

3. 3

   Use the right triangle: the angle is between the line and its projection.

Worked example

1. 1

   $\text{drop }B\text{ to }F$ — $B$ is straight above $F$, so $BF$ is the perpendicular and $F$ is the foot.

2. 2

   $\text{join }F\text{ to }H$ — $H$ is where $BH$ meets the face $EFGH$. The projection is $FH$, the shadow of $BH$.

3. 3

   $F{H}^2=6^2+3^2$ — $FH$ is the diagonal of the rectangle $EFGH$, the longest side of a right triangle with sides 6 and 3.

4. 4

   $F{H}^2=36+9=45$ — Square each side and add.

5. 5

   $\tan \theta =\frac{BF}{FH}=\frac{4}{\sqrt{45}}$ — Triangle $BFH$ has its right angle at $F$. $BF$ is across from $\theta$ (opposite) and $FH$ is next to it (adjacent). Opposite and adjacent mean tan.

6. 6

   $\theta ={\tan }^{-1}(0.5963\dots )$ — We know the ratio and want the angle, so use inverse tan.

7. 7

   $\theta =30.8^{\circ }$ — Round the angle to 1 decimal place.

Worked example

1. 1

   $\sin \theta =\frac{BF}{BH}=\frac{5}{13}$ — $BF$ is opposite $\theta$ and $BH$ is the hypotenuse. You know both, so sin works directly.

2. 2

   $\theta =22.6^{\circ }$ — Inverse sin, rounded to 1 decimal place.

Worked example

1. 1

   $M\text{ drops to }N,N\text{ the middle of }DC$ — Drop $M$ straight down to the base. $M$ is the middle of the top edge, so its foot $N$ is the middle of $DC$.

2. 2

   $DN=\frac{1}{2}\times 8=4$ — $N$ is halfway along $DC$, which is 8 long.

3. 3

   $A{N}^2=4^2+4^2$ — The projection $AN$ is the longest side of triangle $ADN$, with its right angle at $D$. $AD=4$ and $DN=4$.

4. 4

   $A{N}^2=32$ — 16 + 16 = 32.

5. 5

   $\tan \theta =\frac{MN}{AN}=\frac{5}{\sqrt{32}}$ — Triangle $MNA$ has its right angle at $N$. $MN$ is the height 5 (opposite) and $AN$ is the projection (adjacent). Opposite and adjacent mean tan.

6. 6

   $\theta =41.5^{\circ }$ — Inverse tan, rounded to 1 decimal place.

Worked example

1. 1

   $\text{plane}=ABFE$ — The plane is the upright face $ABFE$, so the perpendicular must meet that face, not the base.

2. 2

   $C\text{ drops to }B$ — $BC$ is at right angles to both $AB$ and $BF$, so $BC$ is perpendicular to the face. The foot is $B$.

3. 3

   $\text{join }B\text{ to }E$ — $E$ is where $EC$ meets the face. The projection is $EB$.

4. 4

   $E{B}^2=9^2+4^2$ — $EB$ is the diagonal of the rectangle $ABFE$: the longest side of a right triangle with sides 9 and 4.

5. 5

   $E{B}^2=81+16=97$ — Square each side and add.

6. 6

   $\tan \theta =\frac{BC}{EB}=\frac{6}{\sqrt{97}}$ — Triangle $EBC$ has its right angle at $B$. $BC=6$ is across from $\theta$ (opposite) and $EB$ is adjacent.

7. 7

   $\theta =31.4^{\circ }$ — Inverse tan, rounded to 1 decimal place.

Worked example

1. 1

   $M\text{ is the middle of }AB$ — Triangle $ABC$ is isosceles, so $C$ is straight above the middle of $AB$. Call that point $M$.

2. 2

   $BM=\frac{1}{2}\times 10=5$ — $M$ is halfway along $AB$.

3. 3

   $C{M}^2=13^2-5^2$ — Triangle $CMB$ has its right angle at $M$. $CB$ is the longest side, so subtract.

4. 4

   $C{M}^2=169-25=144,CM=12$ — Take 25 from 169, then take the square root. $CM$ is the height of the prism.

5. 5

   $M{E}^2=5^2+8^2$ — The projection of $CE$ is $ME$. It is the longest side of triangle $MBE$, with its right angle at $B$.

6. 6

   $M{E}^2=89$ — 25 + 64 = 89.

7. 7

   $\tan \theta =\frac{CM}{ME}=\frac{12}{\sqrt{89}}$ — Triangle $CME$ has its right angle at $M$. $CM$ is across from $\theta$ (opposite) and $ME$ is next to it (adjacent).

8. 8

   $\theta =51.8^{\circ }$ — Inverse tan, rounded to 1 decimal place.