3D Trigonometry Worked ExamplesIGCSE Maths 0580

Work in two stages. First find the diagonal across the base, using Pythagoras on the two base edges.

That base diagonal and the vertical height form a right-angled triangle whose hypotenuse is the space diagonal.

Square root.

The angle between a line and a plane is measured between the line and its projection onto that plane — here, the base diagonal.

So the right-angled triangle has the base diagonal ($10$) adjacent and the height ($5$) opposite.

Inverse tangent.

Drop a perpendicular from the apex to the centre of the base. The horizontal distance from the centre to the midpoint of an edge is half the base side.

That distance and the vertical height form a right-angled triangle.

Square root.

The horizontal distance is now from the centre to a corner, which is half the base diagonal.

Halve it.

Apply Pythagoras with the vertical height.

Square root.

The angle between a line and a plane is the angle between the line and its projection onto the plane. $V$ is directly above $O$, so the projection of $VA$ onto the base is $AO$, and the angle is $\theta =$ angle $VAO$ in the right-angled triangle $VOA$.

$AO$ is half the diagonal $AC$ of the square base. Find $AC$ with Pythagoras.

In triangle $VOA$, $VO = 12$ is opposite $\theta$ and $AO$ is adjacent.

(a) The radius, height and slant height form a right-angled triangle.

Square root — a Pythagorean triple.

(b) The height is opposite the base angle and the radius adjacent.

(c) The vertical cross-section is an isosceles triangle with two base angles of $73.74^{\circ}$. The apex angle is what remains of $180^{\circ}$.

(a) $F$ is vertically above $C$, so triangle $ACF$ has a right angle at $C$. First find $AC$, the diagonal of the base, from the right-angled triangle $ABC$.

Now use Pythagoras in the vertical triangle $ACF$.

(b) The angle between a line and a plane is the angle between the line and its projection onto the plane. Dropping a perpendicular from $F$ lands on $C$, so the projection of $AF$ is $AC$ and the angle is $\theta =$ angle $FAC$.

In triangle $ACF$, $CF = 4$ is opposite $\theta$ and $AC = 13$ is adjacent.

(a) Triangle $TPQ$ is vertical with a right angle at $P$. $PT$ is opposite the $25^{\circ}$ angle and $PQ$ is adjacent.

(b) The angle of elevation from $R$ is angle $TRP$, in the vertical triangle $TPR$. First find the horizontal distance $PR$, the diagonal of the field.

Use the unrounded height from (a).

(c) $M$ is the midpoint of $RS$, so $SM = 20\text{ m}$, and $PS = QR = 30\text{ m}$. Triangle $PSM$ is right-angled at $S$.

Now the vertical triangle $TPM$.

(a) $O$ is the midpoint of the base diagonal $AC$. Find $AC$ from the right-angled triangle $ABC$, then halve it.

Triangle $VOA$ is right-angled at $O$, with hypotenuse $VA = 26$. For a 'show that', write every stage.

(b) The projection of $VA$ onto the base is $OA$, so the angle is $VAO$. All three sides of triangle $VOA$ are known; use the two given ones.

(c) The projection of $VM$ is $OM$. $M$ lies on $AB$, so $OM$ runs parallel to $BC$ and is half of $BC$, not half of $AB$.

Similarly, $N$ lies on $BC$, so $ON$ is half of $AB$.

$VM$ rises over the shorter horizontal distance, so it is steeper. Subtract the unrounded angles.

(a) $QR$ lies on the ground, in triangle $PQR$. Two sides and the included angle are known, so use the cosine rule.

Square root, keeping the full value for (b).

(b) The angle of depression is measured down from the horizontal at $T$. Draw the horizontal from $T$ to the point directly above $R$ at the same height as $T$: that makes a vertical right-angled triangle whose horizontal side equals $QR$.

The vertical side is the difference in the heights of the masts.

The vertical side is opposite the angle $d$ and the horizontal side is adjacent.

No lengths are given, but the angle depends only on the shape. Choose a convenient value: let $AB = BC = 1$ unit, so the height $CG = AE = 2$ units.

$G$ is directly above $C$, so the projection of $AG$ onto the base is $AC$ and the angle required is $\theta =$ angle $GAC$, in the right-angled triangle $ACG$. First find $AC$, the diagonal of the square base.

In triangle $ACG$, $CG = 2$ is opposite $\theta$ and $AC = \sqrt{2}$ is adjacent.

Check with a different choice, $AB = 5$ and $AE = 10$: $\tan\theta = \frac{10}{\sqrt{50}} = \sqrt{2}$ again, so the angle is the same. ✓