C6.2; E6.2 - Bearings and linked right triangles

C6.2; E6.2 - Bearings and linked right triangles

A map or a larger shape may hide several right triangles. You will choose a useful triangle, turn bearings into the angle a ratio needs, and carry an unrounded shared length from one triangle into the next.

Selecting a useful right triangle

A route goes 30 m30\,\mathrm{m} east, then 40 m40\,\mathrm{m} north. A straight path joins its start and finish. What information would let you find that path if the northward distance were missing? A right angle plus two sides suggests Pythagoras; a right angle plus an acute angle and one side suggests trigonometry.

Draw the relevant triangle and label only what the question supplies. East-west and north-south directions are perpendicular. Rectangle corners, stated perpendicular lines and previously established angle facts can also justify right angles. A sloping-looking or nearly square corner is not evidence on its own.

Information in one right triangleUseful method
Two side lengths, needing the thirdPythagoras
One acute angle and one side, needing another sideSine, cosine or tangent
Two side lengths, needing an acute angleAn inverse trigonometric ratio

The table is a choice guide, not a rule that forbids other valid methods. For instance, you could find an angle first and then a length, but that would add work when Pythagoras already has both lengths.

[DIAGRAM: asset_name: Selecting a right triangle method; asset_slug: trig_select_triangle; caption: The available information determines the method.; description: Left: a rectangular field has perpendicular sides thirty and forty metres and diagonal d. Right: the same thirty-metre base and a twenty-eight-degree angle to the diagonal are given instead of the height. Both right angles are marked.]
The available information determines the method.

Worked example 1: two perpendicular lengths

Question. From AA, a route goes 30 m30\,\mathrm{m} east to BB, then 40 m40\,\mathrm{m} north to CC. Calculate the straight-line distance ACAC.

Plan. Use Pythagoras in right triangle ABCABC, with ACAC as hypotenuse.

AC2=302+402east and north are perpendicular=900+1600=2500add squared lengthsAC=50positive square root\begin{aligned} AC^2&=30^2+40^2 && \text{east and north are perpendicular}\\ &=900+1600=2500 && \text{add squared lengths}\\ AC&=50 && \text{positive square root} \end{aligned}

Answer. AC=50 mAC=50\,\mathrm{m}.

Check. The lengths are ten times the 3,4,53,4,5 right triangle. The shortcut 5050 is shorter than the 70 m70\,\mathrm{m} route along the two edges.

Now replace the known northward length by an angle of 28∘28^\circ between ABAB and ACAC. One known side remains, so change method.

Worked example 2: an angle instead of a second side

Question. AB=30 mAB=30\,\mathrm{m} is horizontal, BCBC is perpendicular to ABAB, and angle BAC=28∘BAC=28^\circ. Calculate ACAC, correct to 33 significant figures.

Plan. Use cosine because ABAB is adjacent to the given angle and ACAC is the hypotenuse.

cos⁡28∘=30ACadjacent divided by hypotenuseAC=30cos⁡28∘clear the unknown denominator=33.9771…calculate in degrees\begin{aligned} \cos28^\circ&=\frac{30}{AC} && \text{adjacent divided by hypotenuse}\\ AC&=\frac{30}{\cos28^\circ} && \text{clear the unknown denominator}\\ &=33.9771\ldots && \text{calculate in degrees} \end{aligned}

Answer. AC=34.0 mAC=34.0\,\mathrm{m}, correct to 33 significant figures.

Check. Tangent gives BC=30tan⁡28∘BC=30\tan28^\circ. Pythagoras then gives 302+(30tan⁡28∘)2=33.9771…\sqrt{30^2+(30\tan28^\circ)^2}=33.9771\ldots, confirming the distance independently of the cosine rearrangement.

Watch out

AC=302+282AC=30^2+28^2 treats a degree angle as a length and omits the square root. Pythagoras combines three lengths. An angle and one length call for a ratio instead.

Select the two sides relative to the given angle before calculating.

Bearings and angles in a diagram

A bearing measures direction clockwise from north at the starting point. North is 000∘000^\circ, east 090∘090^\circ, south 180∘180^\circ and west 270∘270^\circ. Whole-degree bearings use three digits; a calculated bearing can also include decimal places when requested.

The small angle in a right triangle is not automatically the bearing. Draw north at the starting point and place the destination in its correct quadrant. For a destination west of north, the small angle to north is measured the opposite way from the bearing, so subtract it from 360∘360^\circ.

[DIAGRAM: asset_name: Bearings on either side of north; asset_slug: trig_bearing_quadrants; caption: A small angle to north is not always the clockwise bearing.; description: Two maps show B eight kilometres north and six kilometres east of A, then eight kilometres north and six kilometres west of A. North arrows and the small angle to north are marked. The western case needs three hundred and sixty degrees minus the small angle.]
A small angle to north is not always the clockwise bearing.

Use the east-west distance opposite the small angle to north, and the north-south distance adjacent to it. This is a geometric side choice, not a new trigonometric formula.

Worked example 3: east of north

Question. BB is 8 km8\,\mathrm{km} north and 6 km6\,\mathrm{km} east of AA. Calculate the bearing of BB from AA, correct to 11 decimal place.

Plan. Find the angle east of north using tangent; here it is already clockwise from north.

tan⁡θ=68east distance over north distanceθ=tan⁡−1(68)acute angle from north=36.8698…∘destination is north-east\begin{aligned} \tan\theta&=\frac68 && \text{east distance over north distance}\\ \theta&=\tan^{-1}\left(\frac68\right) && \text{acute angle from north}\\ &=36.8698\ldots^\circ && \text{destination is north-east} \end{aligned}

Answer. The bearing is 036.9∘036.9^\circ, correct to 11 decimal place.

Check. The angle above east is tan⁡−1(8/6)=53.1301…∘\tan^{-1}(8/6)=53.1301\ldots^\circ. Subtracting it from 90∘90^\circ gives 36.8698…∘36.8698\ldots^\circ, and the bearing lies between north and east.

Change only the horizontal direction from east to west. The triangle has the same lengths, but the bearing must now lie between west and north.

Worked example 4: west of north

Question. BB is 8 km8\,\mathrm{km} north and 6 km6\,\mathrm{km} west of AA. Calculate the bearing of BB from AA, correct to 11 decimal place.

Plan. Find the small angle west of north, then complete the clockwise turn.

α=tan⁡−1(68)same small right-triangle angle=36.8698…∘retain the unrounded anglebearing=360∘−αclockwise to north-west=323.1301…∘subtract before rounding\begin{aligned} \alpha&=\tan^{-1}\left(\frac68\right) && \text{same small right-triangle angle}\\ &=36.8698\ldots^\circ && \text{retain the unrounded angle}\\ \text{bearing}&=360^\circ-\alpha && \text{clockwise to north-west}\\ &=323.1301\ldots^\circ && \text{subtract before rounding} \end{aligned}

Answer. The bearing is 323.1∘323.1^\circ, correct to 11 decimal place.

Check. From west towards north, the remaining angle is tan⁡−1(8/6)=53.1301…∘\tan^{-1}(8/6)=53.1301\ldots^\circ. Adding it to west's 270∘270^\circ gives 323.1301…∘323.1301\ldots^\circ.

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