6.02d, 10.05d - The sine rule

6.02d, 10.05d - The sine rule

Find distances across a triangle without a right angle. The formula in 6.02d comes from a shared perpendicular height; 10.05d uses it to find sides and angles, including the cases where two different triangles fit the same measurements.

Why the sine rule works

Two surveyors stand 40 m40\,\mathrm{m} apart at AA and BB. They measure the angles towards a tree at CC: A=48∘A=48^\circ and B=67∘B=67^\circ. Can they find the distance ACAC without reaching the tree? There is no right angle in their triangle, but we can create two right-angled triangles with one perpendicular.

First recall a familiar triangle: sides 55, 535\sqrt3 and 1010 lie opposite angles 30∘30^\circ, 60∘60^\circ and 90∘90^\circ. Dividing each side by the sine of its opposite angle gives 1010 every time. The matching of opposites is the important feature.

Opposite side and angle

In triangle ABCABC, side a=BCa=BC is opposite angle AA, side b=ACb=AC is opposite angle BB, and side c=ABc=AB is opposite angle CC. An opposite side does not touch the vertex of its angle.

Draw a perpendicular of height hh from CC to the line ABAB. The two right-angled triangles express the same height in two ways.

[DIAGRAM: asset_name: A shared height gives the sine rule; asset_slug: sine-shared-height; caption: The same perpendicular height equals both b sin A and a sin B.; description: An acute triangle ABC with AB horizontal, C above the base and D the perpendicular foot. AC is b, BC is a, AB is c and CD is h. A second panel has obtuse angle A and D on the extension of BA, showing that the height remains b sin A.]
The same perpendicular height equals both b sin A and a sin B.

sin⁡A=hbuse triangle ACDsin⁡B=hause triangle BCDbsin⁡A=asin⁡Bboth equal hasin⁡A=bsin⁡Bdivide by both sines\begin{aligned} \sin A&=\frac hb &&\text{use triangle }ACD\\ \sin B&=\frac ha &&\text{use triangle }BCD\\ b\sin A&=a\sin B &&\text{both equal }h\\ \frac a{\sin A}&=\frac b{\sin B} &&\text{divide by both sines} \end{aligned}

For an obtuse angle, the perpendicular can meet an extended side. The small right-triangle angle is 180∘−A180^\circ-A. A reflected right triangle has the same height, giving sin⁡(180∘−A)=sin⁡A\sin(180^\circ-A)=\sin A, so the argument still works. Dropping a height from a different vertex gives the third equal ratio.

Sine rule

For any non-degenerate triangle, with each side opposite its matching angle,

asin⁡A=bsin⁡B=csin⁡C.\frac a{\sin A}=\frac b{\sin B}=\frac c{\sin C}.

Sides are positive and each interior angle lies strictly between 0∘0^\circ and 180∘180^\circ. The reciprocal form, sin⁡Aa=sin⁡Bb=sin⁡Cc\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}, is equivalent.

In the June and November 2027 Higher Tier Formulae Sheet.

Try the relationship on exact values before using calculator decimals.

Worked example 1: three equal ratios

Question. A triangle has A=30∘A=30^\circ, B=60∘B=60^\circ, C=90∘C=90^\circ, with opposite sides a=6a=6, b=63b=6\sqrt3, c=12c=12. Work out the common sine-rule ratio.

Plan. Match each side with its opposite angle, then use the exact sines.

6sin⁡30∘=61/2=1263sin⁡60∘=633/2=1212sin⁡90∘=12\begin{aligned} \frac6{\sin30^\circ}&=\frac6{1/2}=12\\ \frac{6\sqrt3}{\sin60^\circ}&=\frac{6\sqrt3}{\sqrt3/2}=12\\ \frac{12}{\sin90^\circ}&=12 \end{aligned}

Answer. All three ratios are 1212.

Check. Pythagoras gives 62+(63)2=36+108=1226^2+(6\sqrt3)^2=36+108=12^2, confirming that these lengths fit the stated right-angled triangle.

Double every side while keeping the angles fixed. This changes the common ratio, although the trigonometric values stay the same.

Worked example 2: scaling the triangle

Question. The opposite sides for the same three angles are now 1212, 12312\sqrt3 and 2424. Work out the common sine-rule ratio.

Plan. Divide the new sides by the same exact sines.

121/2=241233/2=24241=24\begin{aligned} \frac{12}{1/2}&=24\\ \frac{12\sqrt3}{\sqrt3/2}&=24\\ \frac{24}{1}&=24 \end{aligned}

Answer. The common ratio is 2424.

Check. The triangle is an enlargement with scale factor 22. A ratio consisting of a length divided by a fixed number should also double: 2×12=242\times12=24.

The ratio is equal within one triangle; it is not the same fixed number for every triangle.

Watch out

The expression a/sin⁡Ba/\sin B mismatches the labels. Trace across the triangle from the angle to its opposite side before writing a ratio.

Use one opposite pair to find the common ratio in a fresh triangle.

Finding sides with the sine rule

To find a side, use one complete opposite side–angle pair and the angle opposite the unknown side.

If two angles are known, find the third using the angle sum 180∘180^\circ. Write just the two ratios you need, rather than carrying all three through the calculation. All angles here are in degrees: check the calculator mode.

Rearranging for a side

If aa and AA are a known opposite pair and BB is known,

b=asin⁡Bsin⁡A.b=\frac{a\sin B}{\sin A}.

The labels can be changed to match the triangle. The sine rule is in the 2027 Higher formula sheet; this is its rearrangement.

We can now answer the surveyors' distance question. Remember that ACAC is opposite BB, while the measured baseline ABAB is opposite CC.

Worked example 3: a distance across a triangle

Question. AB=40 mAB=40\,\mathrm{m}, A=48∘A=48^\circ and B=67∘B=67^\circ. Find ACAC to 33 significant figures.

Plan. Find CC, then match ACAC with BB and ABAB with CC.

C=180∘−48∘−67∘=65∘ACsin⁡67∘=40sin⁡65∘AC=40sin⁡67∘sin⁡65∘=40.626589…\begin{aligned} C&=180^\circ-48^\circ-67^\circ=65^\circ\\ \frac{AC}{\sin67^\circ}&=\frac{40}{\sin65^\circ}\\ AC&=\frac{40\sin67^\circ}{\sin65^\circ}\\ &=40.626589\ldots \end{aligned}

Answer. AC=40.6 mAC=40.6\,\mathrm{m} to 33 significant figures.

Check. 67∘67^\circ is slightly larger than 65∘65^\circ, so its opposite side should be slightly longer than 40 m40\,\mathrm{m}. The result has that size.

Change only the measured angle at BB, keeping the baseline and angle at AA. The third angle changes too, so both sines in the calculation must be updated.

Worked example 4: a different sightline

Question. AB=40 mAB=40\,\mathrm{m}, A=48∘A=48^\circ and B=37∘B=37^\circ. Find ACAC to 33 significant figures.

Plan. Find the new angle opposite the baseline before using the sine rule.

C=180∘−48∘−37∘=95∘ACsin⁡37∘=40sin⁡95∘AC=40sin⁡37∘sin⁡95∘=24.164554…\begin{aligned} C&=180^\circ-48^\circ-37^\circ=95^\circ\\ \frac{AC}{\sin37^\circ}&=\frac{40}{\sin95^\circ}\\ AC&=\frac{40\sin37^\circ}{\sin95^\circ}\\ &=24.164554\ldots \end{aligned}

Answer. AC=24.2 mAC=24.2\,\mathrm{m} to 33 significant figures.

Check. The baseline is now opposite the largest angle, 95∘95^\circ, so it should be longer than ACAC. The remaining side is 40sin⁡48∘/sin⁡95∘≈29.8 m40\sin48^\circ/\sin95^\circ\approx29.8\,\mathrm{m}, and 24.2+29.8>4024.2+29.8>40, as a triangle requires.

A sketch helps prevent the label swap that would invert this result.

[DIAGRAM: asset_name: Survey triangles with matched opposite pairs; asset_slug: sine-survey-pairs; caption: AC is opposite B, while the 40 m baseline is opposite C.; description: Two accurately constructed triangles have AB equal to 40 metres and angle A equal to 48 degrees. The first has angle B 67 degrees and C 65 degrees. The second has B 37 degrees and C 95 degrees. AC is highlighted in each and all three vertices are labelled.]
AC is opposite B, while the 40 m baseline is opposite C.

Exam tip

Keep the full calculator value until the final answer. If another side is needed later, use the stored value rather than the rounded displayed answer.

Find the third angle first in the next triangle.

Continue this lesson.

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