E7.5 - The sine rule (Extended)

E7.5 - The sine rule (Extended)

Two sides and an angle can sometimes describe two different triangles. The sine rule lets you calculate in triangles without a right angle, but you must match each side to its opposite angle and check which triangles are possible. You will find sides, find acute or obtuse angles and resolve the ambiguous case.

Paired sides and angles

Suppose a triangle has A=40∘A=40^\circ, B=65∘B=65^\circ and side a=8 cma=8\,\mathrm{cm} opposite AA. How long is the side opposite BB? There is no right angle, so a single right-triangle ratio cannot link the given sides directly.

Opposite side and angle

In triangle ABCABC, side a=BCa=BC is opposite angle AA, side b=CAb=CA is opposite angle BB, and side c=ABc=AB is opposite angle CC. Opposite means the side does not touch the angle's vertex.

A perpendicular height from CC to the line ABAB can be written as bsin⁡Ab\sin A and asin⁡Ba\sin B. Equating the same height gives bsin⁡A=asin⁡Bb\sin A=a\sin B. Rearranging produces the sine rule. This relationship extends to obtuse triangles by using the extension of a side for the perpendicular.

Sine rule

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

Each numerator is paired with its opposite angle. The rule applies to any triangle. Supplied in the Extended formula list on Papers 2 and 4.

Choose two matching fractions. For a missing side, you need one complete opposite pair and the angle opposite the required side. If two angles are known but the useful third angle is missing, first subtract the known angles from 180∘180^\circ.

Worked example 1: finding a side

Question. Triangle ABCABC has A=40∘A=40^\circ, B=65∘B=65^\circ and a=8 cma=8\,\mathrm{cm}. Calculate bb.

Plan. Pair bb with 65∘65^\circ and 8 with 40∘40^\circ, then multiply to isolate bb.

bsin⁡65∘=8sin⁡40∘b=8sin⁡65∘sin⁡40∘b=11.2797…\begin{aligned} \frac{b}{\sin65^\circ}&=\frac8{\sin40^\circ}\\ b&=\frac{8\sin65^\circ}{\sin40^\circ}\\ b&=11.2797\ldots \end{aligned}

Answer. b=11.3 cmb=11.3\,\mathrm{cm}, to 3 significant figures.

Check. 65∘>40∘65^\circ>40^\circ, so its opposite side must be longer than 8. Using the unrounded answer gives the same value of b/sin⁡65∘b/\sin65^\circ as 8/sin⁡40∘8/\sin40^\circ.

Change only angle BB to an obtuse angle. The same paired calculation still works.

Worked example 2: a side opposite an obtuse angle

Question. Triangle ABCABC has A=40∘A=40^\circ, B=95∘B=95^\circ and a=8 cma=8\,\mathrm{cm}. Calculate bb.

Plan. Keep the opposite pairings and enter the whole expression in degree mode.

bsin⁡95∘=8sin⁡40∘b=8sin⁡95∘sin⁡40∘b=12.3984…\begin{aligned} \frac{b}{\sin95^\circ}&=\frac8{\sin40^\circ}\\ b&=\frac{8\sin95^\circ}{\sin40^\circ}\\ b&=12.3984\ldots \end{aligned}

Answer. b=12.4 cmb=12.4\,\mathrm{cm}, to 3 significant figures.

Check. The third angle is 45∘45^\circ, so BB is the largest angle and bb should be the longest side. Its calculated value is larger than 8; the third side from the sine rule is about 8.80 cm8.80\,\mathrm{cm}, also shorter.

The size comparison is a check, not a way to calculate the missing side.

Watch out

Write the opposite pairs before substituting. b/sin⁡40∘=8/sin⁡95∘b/\sin40^\circ=8/\sin95^\circ swaps the pairings and gives the wrong length. Keep unrounded values until the final answer.

Try the same method with new data.

Finding angles

To find an angle, invert both fractions so that its sine is on top: sin⁡B/b=sin⁡A/a\sin B/b=\sin A/a. Rearranging gives sin⁡B=bsin⁡A/a\sin B=b\sin A/a. The inverse sine key gives the acute answer when the sine is strictly between 0 and 1, but a triangle angle may also be obtuse.

Supplementary sine values

For 0∘<θ<90∘0^\circ<\theta<90^\circ, the angles θ\theta and 180∘−θ180^\circ-\theta have the same sine. Reflecting a sloping side from one side of a perpendicular to the other preserves its height and length, so the height-to-length ratio stays the same.

Thus sin⁡50∘=sin⁡130∘\sin50^\circ=\sin130^\circ. The second angle is found by subtracting from 180∘180^\circ, not 360∘360^\circ. Every interior triangle angle must be strictly between 0∘0^\circ and 180∘180^\circ.

Worked example 3: an acute angle

Question. Triangle ABCABC has a=9 cma=9\,\mathrm{cm}, b=12 cmb=12\,\mathrm{cm} and A=35∘A=35^\circ. Angle BB is acute. Calculate BB.

Plan. Calculate sin⁡B\sin B, then choose the acute inverse-sine result.

sin⁡B12=sin⁡35∘9sin⁡B=12sin⁡35∘9=0.764768…B=sin⁡−1(0.764768…)=49.8864…∘\begin{aligned} \frac{\sin B}{12}&=\frac{\sin35^\circ}{9}\\ \sin B&=\frac{12\sin35^\circ}{9}=0.764768\ldots\\ B&=\sin^{-1}(0.764768\ldots)=49.8864\ldots^\circ \end{aligned}

Answer. B=49.9∘B=49.9^\circ, to 1 decimal place.

Check. C=95.1135…∘C=95.1135\ldots^\circ is positive. Also b>ab>a and B>AB>A, agreeing with the order of opposite sides and angles.

Keep the same data but require BB to be obtuse. The calculator's principal answer must now be supplemented.

Worked example 4: an obtuse angle

Question. Triangle ABCABC has a=9 cma=9\,\mathrm{cm}, b=12 cmb=12\,\mathrm{cm} and A=35∘A=35^\circ. Angle BB is obtuse. Calculate BB.

Plan. Find the acute angle with the required sine, then subtract it from 180∘180^\circ.

sin⁡B=12sin⁡35∘9B=180∘−49.8864…∘=130.1135…∘\begin{aligned} \sin B&=\frac{12\sin35^\circ}{9}\\ B&=180^\circ-49.8864\ldots^\circ\\ &=130.1135\ldots^\circ \end{aligned}

Answer. B=130.1∘B=130.1^\circ, to 1 decimal place.

Check. C=180∘−35∘−130.1135…∘=14.8864…∘>0C=180^\circ-35^\circ-130.1135\ldots^\circ=14.8864\ldots^\circ>0. Substituting the unrounded BB gives sin⁡B=0.764768…\sin B=0.764768\ldots, as required.

A diagram can suggest which angle is intended, but an explicit statement such as “obtuse” settles it. If no such restriction is given, test both possibilities, as in the next part.

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