HG22 - Sine and cosine rules

HG22 - Sine and cosine rules

Find sides and angles in triangles which need not contain a right angle. The sine rule links opposite side–angle pairs; the cosine rule uses two sides and their included angle, or all three sides. This is Higher content.

Sine rule

A triangle has angles 40∘40^\circ and 65∘65^\circ, with side 8 cm8\,\mathrm{cm} opposite 40∘40^\circ. How can you find the side opposite 65∘65^\circ without a right angle?

Label vertices A,B,CA,B,C and their opposite sides a,b,ca,b,c: a=BCa=BC, b=CAb=CA, c=ABc=AB. The letters are a pairing convention, not a requirement that cc is a hypotenuse. Only a right-angled triangle has a hypotenuse.

[DIAGRAM: asset_name: Opposite pairs and obtuse angles; asset_slug: general-triangle-pairs; caption: Match each side to the angle opposite it; reflected unit-circle points explain supplementary-angle signs.; description: An exact scalene triangle labelled A B C with opposite sides a b c, beside a unit semicircle showing 60 and 120 degree radii with equal positive heights and opposite horizontal coordinates.]
Match each side to the angle opposite it

For an acute triangle, dropping a perpendicular from CC to ABAB gives the same height as bsin⁡Ab\sin A and asin⁡Ba\sin B. Thus bsin⁡A=asin⁡Bb\sin A=a\sin B, which rearranges to a/sin⁡A=b/sin⁡Ba/\sin A=b/\sin B. Repeating with another height brings in the third pair.

For angles between 90∘90^\circ and 180∘180^\circ, use the unit-circle interpretation: sine is the vertical coordinate, cosine the horizontal coordinate. Reflecting a radius across the vertical axis gives the supplementary angle 180∘−θ180^\circ-\theta. Its height is unchanged and its horizontal coordinate changes sign:

sin⁡(180∘−θ)=sin⁡θ,cos⁡(180∘−θ)=−cos⁡θ.\sin(180^\circ-\theta)=\sin\theta,\qquad \cos(180^\circ-\theta)=-\cos\theta.

So a triangle's obtuse angle has positive sine and negative cosine. An outside altitude and the equal supplementary sines extend the sine rule to obtuse triangles too.

Sine rule

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

Use two opposite pairs, including one complete known pair. Angles are interior angles of a nondegenerate triangle: positive and summing to 180∘180^\circ; side lengths are positive. The rule is supplied on the current Higher additional formula sheet.

A complete opposite pair provides the common ratio; use it to calculate the missing member of a second pair.

Worked example 1: finding a side

Question. In triangle ABCABC, A=40∘A=40^\circ, B=65∘B=65^\circ, and a=8 cma=8\,\mathrm{cm}. Find bb to 33 significant figures.

Plan. Pair 88 with 40∘40^\circ, and bb with 65∘65^\circ, then isolate bb.

bsin⁡65∘=8sin⁡40∘,b=8sin⁡65∘sin⁡40∘=11.279…\frac{b}{\sin65^\circ}=\frac8{\sin40^\circ}, \qquad b=\frac{8\sin65^\circ}{\sin40^\circ} =11.279\ldots

Answer. b=11.3 cmb=11.3\,\mathrm{cm} to 33 significant figures.

Check. The remaining angle is 75∘75^\circ. Its side is c=8sin⁡75∘/sin⁡40∘=12.021…c=8\sin75^\circ/\sin40^\circ=12.021\ldots. The ordering 8<11.279…<12.021…8<11.279\ldots<12.021\ldots matches 40∘<65∘<75∘40^\circ<65^\circ<75^\circ, and the two shorter sides sum to more than the longest.

Change to finding an angle. Inverse sine returns the acute candidate; an obtuse supplement may need checking, as the third part explains.

Worked example 2: finding a specified acute angle

Question. A=50∘A=50^\circ, a=10 cma=10\,\mathrm{cm}, b=7 cmb=7\,\mathrm{cm}, and BB is acute. Find BB to 11 decimal place.

Plan. Isolate sin⁡B\sin B, then use inverse sine in degree mode.

sin⁡B7=sin⁡50∘10,sin⁡B=7sin⁡50∘10,B=sin⁡−1 ⁣(7sin⁡50∘10)=32.427…∘.\frac{\sin B}{7}=\frac{\sin50^\circ}{10}, \quad \sin B=\frac{7\sin50^\circ}{10}, \quad B=\sin^{-1}\!\left(\frac{7\sin50^\circ}{10}\right) =32.427\ldots^\circ.

Answer. B=32.4∘B=32.4^\circ.

Check. b<ab<a requires B<AB<A, which holds. The alternative 147.572…∘147.572\ldots^\circ would already give A+B>180∘A+B>180^\circ, so it cannot form this triangle even without the stated acute condition.

The side ordering and angle sum give geometric checks on an inverse-sine result.

Watch out

Do not pair adjacent sides and angles. Write the opposite pairings before substituting, and use the unrounded sine value inside inverse sine.

Use a fresh complete pair.

Cosine rule

If no complete opposite pair is known, the sine rule may not start the problem. SAS means two sides and their included angle; SSS means all three sides. The cosine rule handles both.

The included angle lies between the two known sides. In a2=b2+c2−2bccos⁡Aa^2=b^2+c^2-2bc\cos A, AA is between bb and cc, and aa lies opposite AA. You may relabel the formula to fit any vertex.

Cosine rule: side or angle

a2=b2+c2−2bccos⁡Aa^2=b^2+c^2-2bc\cos A cos⁡A=b2+c2−a22bc.\cos A=\frac{b^2+c^2-a^2}{2bc}.

Take the positive square root for a length. Use inverse cosine for an angle in 0∘<A<180∘0^\circ<A<180^\circ. The side formula is supplied on the Higher additional sheet; the angle form is its rearrangement.

The correction term explains the link to Pythagoras. At 90∘90^\circ, cosine is zero. For an acute included angle, subtract a positive quantity; for an obtuse angle, its cosine is negative, so the correction increases the opposite side's square.

Worked example 3: SAS to a side

Question. Two sides are 7 cm7\,\mathrm{cm} and 10 cm10\,\mathrm{cm}, with included angle 60∘60^\circ. Find the third side aa to 33 significant figures.

Plan. Put the unknown side opposite the known included angle.

a2=72+102−2(7)(10)cos⁡60∘=149−70=79.a^2=7^2+10^2-2(7)(10)\cos60^\circ =149-70=79.

Answer. a=79=8.89 cma=\sqrt{79}=8.89\,\mathrm{cm} to 33 significant figures.

Check. Place the 10 cm10\,\mathrm{cm} side horizontally. The 7 cm7\,\mathrm{cm} side projects 7cos⁡60∘=3.5 cm7\cos60^\circ=3.5\,\mathrm{cm} along it and 7sin⁡60∘=73/2 cm7\sin60^\circ=7\sqrt3/2\,\mathrm{cm} upwards. Pythagoras gives a2=(10−3.5)2+(73/2)2=79a^2=(10-3.5)^2+(7\sqrt3/2)^2=79.

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