PT2 - The sine and cosine rules

Find missing lengths, angles and areas in triangles without a right angle. You will choose a rule from the information given, match opposite sides and angles correctly, and keep enough accuracy for a final answer.

The sine rule and opposite pairs

Two triangles have sides $8\,\mathrm{cm}$ and $11\,\mathrm{cm}$ , but their included angles are $40^\circ$ and $140^\circ$ . Can they have the same area even though their third sides are different? We will settle that puzzle after finding ways to work with triangles of any shape.

You already know

The angles in a triangle total $180^\circ$ . Sine is positive for every angle strictly between $0^\circ$ and $180^\circ$ , including obtuse angles. Use degree mode and round only at the end.

In triangle $ABC$ , use lower-case $a$ for the side opposite $A$ , $b$ for the side opposite $B$ , and $c$ for the side opposite $C$ . Thus $a=BC$ , not $AB$ . “Opposite” means the side that does not touch the angle's vertex.

The sine rule works because one altitude can be expressed in two ways. Taking $AB=c$ as the base, the height from $C$ is both $b\sin A$ and $a\sin B$ , so $b\sin A=a\sin B$ . Rearranging gives equal side-to-sine ratios. An obtuse base angle uses an altitude outside the triangle, with the same sine as its acute supplement, so the rule still holds.

Sine rule

\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}

This applies to every non-degenerate triangle, with each side paired with its opposite angle. The reciprocal form is equally valid.
On the formulae sheet.

Look for a complete opposite pair and one other side or angle. With two angles given, find the third first if necessary. When the unknown is an angle, use the reciprocal form to put its sine on top.

Worked example 1: finding a side

Question. In triangle $ABC$ , $a=9\,\mathrm{cm}$ , $A=42^\circ$ and $B=68^\circ$ . Find $b$ , correct to 3 significant figures.

Plan. Pair $a$ with $A$ , equate this ratio to the one containing $b$ , and rearrange before calculating.

\begin{aligned} \frac{b}{\sin68^\circ}&=\frac9{\sin42^\circ} &&\text{opposite pairs}\\ b&=\frac{9\sin68^\circ}{\sin42^\circ} &&\text{multiply by }\sin68^\circ\\ &=12.4708\ldots &&\text{evaluate in degrees} \end{aligned}

Why choose sine rule? One complete opposite pair, $(9, 42^\circ)$ , is already known; there is no need to find the third side.

Answer. $b=12.5\,\mathrm{cm}$ , to 3 significant figures.

Check. $68^\circ>42^\circ$ , so its opposite side should be longer: $12.5>9$ .

The next example changes the unknown from a side to an angle. The ratio is the same, but its reciprocal form makes the rearrangement shorter.

Worked example 2: finding an angle

Question. In triangle $ABC$ , $a=9\,\mathrm{cm}$ , $A=42^\circ$ and $b=6\,\mathrm{cm}$ . Find $B$ , correct to 1 decimal place.

Plan. Find $\sin B$ , take the inverse sine, and check that the angle fits a triangle.

\begin{aligned} \frac{\sin B}{6}&=\frac{\sin42^\circ}{9} &&\text{reciprocal sine rule}\\ \sin B&=\frac{6\sin42^\circ}{9}=0.44608\ldots &&\text{multiply by }6\\ B_1&=\sin^{-1}(0.44608\ldots)=26.4929\ldots^\circ &&\text{principal angle}\\ B_2&=180^\circ-B_1=153.5070\ldots^\circ &&\text{same sine}\\ 42^\circ+B_2&>180^\circ &&\text{reject the supplement} \end{aligned}

Why check the supplement? An acute angle and its obtuse supplement share the same sine. Here the second candidate leaves no positive third angle; the next lesson explores cases where it survives.

Answer. $B=26.5^\circ$ , to 1 decimal place.

Check. Side $b$ is shorter than $a$ , so $B$ must be smaller than $42^\circ$ , as found.

Before pressing a calculator key, trace the pairs on your sketch.

Watch out

Wrong: $a/\sin B=b/\sin A$ . This crosses the pairs. Right: $a/\sin A=b/\sin B$ . Labelling the opposite side beside each angle prevents the error.

Now find an angle from a different complete pair, checking the supplementary candidate too.

The cosine rule for a missing length

If you know two sides and their included angle, you do not yet have an opposite pair for the sine rule; the cosine rule gives the third side directly.

Place $A$ at the origin and $AB=c$ along the horizontal axis. Then $C$ has coordinates $(b\cos A, b\sin A)$ . The horizontal and vertical separations of $B$ and $C$ give

a^2=(c-b\cos A)^2+(b\sin A)^2=b^2+c^2-2bc\cos A

The last simplification uses the fact that the squared coordinates of a unit-circle point add to $1$ . We will study that identity in detail shortly. This coordinate argument also works when $A$ is obtuse and $b\cos A$ is negative.

Cosine rule for a side

a^2=b^2+c^2-2bc\cos A

In any non-degenerate triangle, $A$ is the angle included between sides $b$ and $c$ , and $a$ is opposite $A$ . Use the positive square root to find a length.
On the formulae sheet.

When $A=90^\circ$ , its cosine is zero and this becomes Pythagoras' theorem. An acute angle has positive cosine, so $a^2<b^2+c^2$ ; an obtuse angle has negative cosine, so $a^2>b^2+c^2$ . That sign gives a quick check.

Worked example 3: an acute included angle

Question. Two sides of a triangle are $7\,\mathrm{cm}$ and $10\,\mathrm{cm}$ , with included angle $55^\circ$ . Find the third side $a$ , correct to 3 significant figures.

Plan. Substitute the two sides and the included angle into cosine rule, then take the positive square root.

\begin{aligned} a^2&=7^2+10^2-2(7)(10)\cos55^\circ &&\text{cosine rule}\\ &=68.6992\ldots &&\text{evaluate the full expression}\\ a&=\sqrt{68.6992\ldots}=8.28850\ldots &&\text{positive length} \end{aligned}

Why use the included angle? The rule connects the unknown opposite side to the two sides touching the known angle.

Answer. $a=8.29\,\mathrm{cm}$ .

Check. The triangle inequality gives $3<a<17$ , and $a^2<149$ because the angle is acute.

The next example changes only the included angle to its obtuse supplement. The cosine changes sign, making the opposite side longer.

Worked example 4: an obtuse included angle

Question. Two sides of a triangle are $7\,\mathrm{cm}$ and $10\,\mathrm{cm}$ , with included angle $125^\circ$ . Find the third side $a$ , correct to 3 significant figures.

Plan. Use the same cosine rule, keeping the negative cosine inside the subtraction.

\begin{aligned} a^2&=7^2+10^2-2(7)(10)\cos125^\circ &&\text{cosine rule}\\ &=149-140(-0.57357\ldots) &&\text{cosine is negative}\\ &=229.3007\ldots &&\text{subtract a negative}\\ a&=15.1426\ldots &&\text{positive square root} \end{aligned}

Why keep the sign visible? The correction term increases the squared length for an obtuse angle; removing the negative would produce the acute triangle instead.

Answer. $a=15.1\,\mathrm{cm}$ .

Check. $\sqrt{149}<a<17$ , and the side opposite $125^\circ$ is the longest.

Write the whole substitution before evaluating it; it makes the grouping clear.

Watch out

Wrong: $a^2=(b^2+c^2-2bc)\cos A$ . Only the term $2bc$ is multiplied by $\cos A$ . Right: $a^2=b^2+c^2-2bc\cos A$ .

Try another obtuse triangle, retaining the unrounded value until the square root is taken.

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Algebra

Enumeration (counting and probability)

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Pythagoras' theorem and trigonometry

Calculus

Exponentials and logarithms

Numerical methods

PT2 - The sine and cosine rules

The sine rule and opposite pairs

The cosine rule for a missing length

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