E6.5 - Sine and cosine rules

E6.5 - Sine and cosine rules

A triangle need not contain a right angle for its sides and angles to be connected. You will choose between the sine and cosine rules, calculate with obtuse angles, and decide when the same given information describes two different triangles. This lesson is for Extended.

Using the sine rule

Can two triangles have two equal given sides and the same given angle, yet still be different? They can if the angle is not between those sides. First learn to match a side to its opposite angle; that pairing will explain both the method and its possible ambiguity.

In triangle ABCABC, lower-case aa is the side opposite angle AA, bb is opposite BB, and cc is opposite CC. A side is opposite the vertex it does not touch.

[DIAGRAM: asset_name: Opposite pairs and perpendicular heights; asset_slug: triangle_rules; caption: Match each lower-case side with its opposite capital angle.; description: An acute labelled triangle ABC shows sides a opposite A, b opposite B and c opposite C, with a perpendicular height h to AB. A second panel reflects one radius-like side across the vertical through its base, showing equal heights for an acute angle theta and its obtuse supplement, while the horizontal projection changes sign.]
Match each lower-case side with its opposite capital angle.

For an acute triangle, dropping height hh from CC gives h=bsin⁡A=asin⁡Bh=b\sin A=a\sin B. Dividing gives sin⁡A/a=sin⁡B/b\sin A/a=\sin B/b, and the third pair follows by choosing another height.

An obtuse angle needs a small extension of the acute ratios. A side of fixed length makes the same positive height at θ\theta and 180∘−θ180^\circ-\theta, as the reflection in the diagram shows. We therefore use

sin⁡(180∘−θ)=sin⁡θfor 0∘<θ<90∘.\sin(180^\circ-\theta)=\sin\theta \quad\text{for }0^\circ<\theta<90^\circ.

The same height argument then works when a perpendicular meets an extended side.

Sine rule

For any non-degenerate triangle,

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

Each side must stay paired with its opposite angle. Every angle is strictly between 0∘0^\circ and 180∘180^\circ.

Given in the List of formulas.

Use the sine rule when you have a complete opposite side-angle pair and another side or angle. With two known angles, find the third using the angle sum if needed. With a known sine, an inverse-sine key gives the acute candidate first; check its supplement too.

Worked example 1: finding a side

Question. In triangle ABCABC, A=48∘A=48^\circ, B=67∘B=67^\circ, and a=12 cma=12\,\mathrm{cm}. Calculate bb.

Plan. Pair bb with 67∘67^\circ and the known side 1212 with 48∘48^\circ. Use degree mode.

bsin⁡67∘=12sin⁡48∘matching opposite pairsb=12sin⁡67∘sin⁡48∘=14.8639…\begin{aligned} \frac{b}{\sin67^\circ}&=\frac{12}{\sin48^\circ} && \text{matching opposite pairs}\\ b&=\frac{12\sin67^\circ}{\sin48^\circ}\\ &=14.8639\ldots \end{aligned}

Answer. b=14.9 cmb=14.9\,\mathrm{cm}, to 33 significant figures.

Check. C=65∘C=65^\circ, so BB is the largest angle. Its opposite side bb should exceed aa, which it does. Substituting the unrounded bb gives both sine-rule ratios 16.1475… cm16.1475\ldots\,\mathrm{cm}.

Now the unknown is an angle rather than its opposite side. The pairing remains the same, but the final operation changes.

Worked example 2: finding an angle

Question. In triangle ABCABC, A=48∘A=48^\circ, a=12 cma=12\,\mathrm{cm}, and b=10 cmb=10\,\mathrm{cm}. Calculate BB.

Plan. Isolate sin⁡B\sin B, then test both angle candidates against the angle sum.

sin⁡B10=sin⁡48∘12sin⁡B=10sin⁡48∘12B1=38.2641…∘inverse sine in degree modeB2=180∘−B1=141.7358…∘same sine\begin{aligned} \frac{\sin B}{10}&=\frac{\sin48^\circ}{12}\\ \sin B&=\frac{10\sin48^\circ}{12}\\ B_1&=38.2641\ldots^\circ && \text{inverse sine in degree mode}\\ B_2&=180^\circ-B_1=141.7358\ldots^\circ && \text{same sine} \end{aligned}

The second candidate fails: 48∘+141.7358…∘>180∘48^\circ+141.7358\ldots^\circ>180^\circ.

Answer. B=38.3∘B=38.3^\circ, to 11 decimal place.

Check. Since b<ab<a, its opposite angle must be smaller than A=48∘A=48^\circ. The accepted angle is smaller; the rejected candidate is not.

Watch out

asin⁡B=bsin⁡A\frac{a}{\sin B}=\frac{b}{\sin A} crosses the pairs. Label opposite sides before substituting. Also, sin⁡−1\sin^{-1} here means inverse sine, not the reciprocal 1/sin⁡1/\sin.

Using the cosine rule

The sine rule cannot start if you know two sides and only the angle between them: no opposite pair is complete. The cosine rule uses exactly that information. It can also find an angle when all three sides are known.

For an acute angle AA, the horizontal projection of side bb onto side cc is q=bcos⁡Aq=b\cos A, and its perpendicular height is h=bsin⁡Ah=b\sin A. Pythagoras gives

a2=(c−q)2+h2=c2−2cq+(q2+h2)=c2−2bccos⁡A+b2since q2+h2=b2.\begin{aligned} a^2&=(c-q)^2+h^2\\ &=c^2-2cq+(q^2+h^2)\\ &=c^2-2bc\cos A+b^2 && \text{since }q^2+h^2=b^2. \end{aligned}

For an obtuse AA, the projection points in the opposite direction, so its signed value is negative. If A=180∘−θA=180^\circ-\theta, reflection gives cos⁡A=−cos⁡θ\cos A=-\cos\theta. The same expansion works with this negative projection. In particular, an obtuse angle has negative cosine.

Cosine rule

a2=b2+c2−2bccos⁡Aa^2=b^2+c^2-2bc\cos A

Here AA is the included angle between the known sides bb and cc, opposite aa. Equivalently,

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

Use positive side lengths of a genuine triangle; for three supplied sides, the two shorter sides must add to more than the longest.

Given in the List of formulas.

The supplied version can be relabelled: the squared side alone must always be opposite the angle used. At 90∘90^\circ, cosine is zero and the rule reduces to Pythagoras.

Worked example 3: two sides and their included angle

Question. Two sides of a triangle are 7 cm7\,\mathrm{cm} and 9 cm9\,\mathrm{cm}. Their included angle is 60∘60^\circ. Find the third side.

Plan. Use the cosine rule for the side opposite 60∘60^\circ, then take the positive square root.

a2=72+92−2(7)(9)cos⁡60∘=49+81−126(12)exact cosine=67a=67length is positive\begin{aligned} a^2&=7^2+9^2-2(7)(9)\cos60^\circ\\ &=49+81-126\left(\frac12\right) && \text{exact cosine}\\ &=67\\ a&=\sqrt{67} && \text{length is positive} \end{aligned}

Answer. 67 cm\sqrt{67}\,\mathrm{cm}, or 8.19 cm8.19\,\mathrm{cm} to 33 significant figures.

Check. Drop a perpendicular from the endpoint of side 77 onto side 99. The projection is 7cos⁡60∘=3.57\cos60^\circ=3.5, the height is 7sin⁡60∘=73/27\sin60^\circ=7\sqrt3/2, and the remaining base is 9−3.5=5.59-3.5=5.5. Pythagoras independently gives a2=5.52+(73/2)2=30.25+36.75=67a^2=5.5^2+(7\sqrt3/2)^2=30.25+36.75=67. The triangle inequalities also hold: 2<67<162<\sqrt{67}<16.

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