PT6 - Solving problems in two dimensions

Combining a height with a sloping length

Will a $25\,\mathrm{m}$ cable reach the top of a mast from an anchor $15\,\mathrm{m}$ from its base? An observer $24\,\mathrm{m}$ from the base sees the top at elevation $38^\circ$, with eyes $1.6\,\mathrm{m}$ above level ground. There is enough information, but it belongs to two different right-angled triangles.

A drawing translates the words into geometry. An angle of elevation is measured upwards from the observer's horizontal line, not from the ground beneath their feet. An angle of depression is measured downwards from a horizontal line. Parallel horizontals make a depression angle equal to the corresponding elevation angle at the other end of the same sightline.

[DIAGRAM: asset_name: Two triangles for a mast and its cable; asset_slug: two-d-height-cable; caption: The observation triangle finds height above eye level; the cable triangle uses the full mast height; description: A vertical mast with top T and base B stands on horizontal ground. Observer E is 24 metres to the left with eye height 1.6 metres. A dashed horizontal from E meets the mast at F and the elevation angle between EF and ET is 38 degrees. A ground anchor A is 15 metres to the right of B. The blue cable AT and full mast BT form a second right-angled triangle. Right-angle marks are at F and B.]

If the vertical difference is $v$ and horizontal separation is $d$, tangent gives $v=d\tan\theta$. Add the observer's height to obtain the full mast height. Pythagoras then gives the cable length from the full vertical and horizontal separations. The model treats the cable as straight and the ground as level.

Name the height you have just found before carrying it into the next triangle. That small habit prevents the eye-level height being mistaken for the full height.

The next example changes only the observer's height above the ground. Their horizontal distance and measured elevation remain the same, so the inferred mast is taller.

A perspective sketch does not prove a right angle; the model's vertical mast and horizontal ground do. Mark those facts explicitly.

Now try the complete chain with new measurements.

Finding a height from two lines of sight

Two observations can reveal both a height and an unknown distance when you cannot walk to the base.

Suppose two points $P$ and $Q$ lie on the same horizontal line as a tower base $B$, with $Q$ between $P$ and $B$. The angles of elevation of top $T$ are $\alpha$ from $P$ and the larger angle $\beta$ from $Q$. Write $QB=x$, $PQ=d$ and $BT=h$. The farther horizontal distance is $x+d$, not $d$.

[DIAGRAM: asset_name: Two observations share one height; asset_slug: two-d-two-sightlines; caption: The two right-angled triangles share height h and have horizontal distances x and x plus d; description: P, Q and B lie in that order on a horizontal baseline, with a vertical tower BT at B. PQ is marked d and QB is marked x. Sightlines PT and QT form angles alpha and beta with the rightward horizontal, beta greater than alpha. The full height BT is h and a right-angle mark appears at B. The larger observation triangle contains the smaller.]

Write one tangent equation for each triangle, then equate the two expressions for the same $h$. This derives the distance from the diagram; it does not need another memorised formula. The resulting two-observation relationship is not on the formulae sheet: derive it.

The next example changes only the nearer elevation to $50^\circ$. A steeper near sightline changes both inferred unknowns.

If the observers stand on opposite sides of the tower, the distance between them is a sum of the two distances to the base. Draw that layout afresh instead of reusing $x+d$. If their eyes are at the same non-zero height, both equations find the height above that common eye level; add the eye height afterwards.

Set up both triangles for a different pair of observations.

Turning bearings into triangle angles

Bearings describe directions from north; a triangle rule needs the interior angle between its own sides.

Draw parallel north lines at each location. Reversing a direction changes its bearing by $180^\circ$, reduced to the range $0^\circ\le b<360^\circ$. This bearing relationship is not on the formulae sheet: learn it. If a journey goes from $A$ to $B$, then to $C$, the angle at $B$ is between $BA$ and $BC$, so reverse the first bearing before comparing directions.