Mt3, Mt4 - Triangle area, sine rule and cosine rule | OCR A-Level Mathematics B (MEI)

Choose and use trigonometric rules for triangles that need not contain a right angle, derive their area from perpendicular height, and translate bearings into the angles a calculation needs.

Area from two sides and their angle

Can the same two side lengths give equal-area triangles with very different shapes? Take sides $6\,\mathrm{cm}$ and $10\,\mathrm{cm}$ . An included angle of $30^\circ$ gives area $15\,\mathrm{cm}^2$ ; so does $150^\circ$ . The shared height is the clue, and it also tells us which angle belongs in the area formula.

You already know

A triangle has area half base times perpendicular height. Right-triangle sine gives height $L\sin\theta$ for a side $L$ inclined at $\theta$ ; for example, a $5\,\mathrm m$ ladder at $30^\circ$ reaches height $2.5\,\mathrm m$ .

Use side $a$ opposite angle $A$ , side $b$ opposite $B$ , and side $c$ opposite $C$ . The angle between $a$ and $b$ is $C$ . Taking $b$ as base, the height is $a\sin C$ , giving the result below. With an obtuse angle, extend the base to draw the height; $\sin(180^\circ-C)=\sin C$ preserves its positive length.

Area from an included angle

\Delta=\frac12ab\sin C=\frac12bc\sin A=\frac12ca\sin B.

Use the angle included between the two chosen positive sides, with interior angles strictly between $0^\circ$ and $180^\circ$ .

Not in the formula sheet: learn it.

There is a useful surprise hiding here: fixing two sides does not make area increase all the way to a straight angle. It rises to a maximum at $90^\circ$ , then falls again.

Worked example 1: a height inside the triangle

Question. Determine the exact area for sides $6\,\mathrm{cm}$ and $10\,\mathrm{cm}$ with included angle $30^\circ$ .

Plan. Substitute the two sides and the angle between them into the area rule.

\begin{aligned} \Delta&=\frac12(6)(10)\sin30^\circ &&\text{included-angle rule}\\ &=30\left(\frac12\right)=15 &&\text{exact sine} \end{aligned}

Why use these two sides together? They meet at the given angle, so their product and that sine give the perpendicular-height area.

Answer. $15\,\mathrm{cm}^2$ .

Check. On base $10$ , the height is $6\sin30^\circ=3$ , so half base times height is $15$ .

Change only the included angle, from $30^\circ$ to its supplement $150^\circ$ . The height now lies outside the triangle, but its length is unchanged.

Worked example 2: an external height

Question. Determine the exact area for the same two sides and included angle $150^\circ$ .

Plan. Use the obtuse sine value directly.

\begin{aligned} \Delta&=\frac12(6)(10)\sin150^\circ &&\text{same area rule}\\ &=30\left(\frac12\right)=15 &&\text{supplementary angles share sine} \end{aligned}

Why not change the sign? The height is positive, and sine is positive for every triangle's interior angle.

Answer. $15\,\mathrm{cm}^2$ , the same area as before.

Check. Both areas are below $30\,\mathrm{cm}^2$ , the right-angle maximum for these side lengths.

The shapes differ, but their equal heights answer the opening puzzle.

Watch out

Wrong: use $\tfrac12ab\sin A$ merely because the letter $a$ appears. Right: $A$ is opposite $a$ ; the angle between $a$ and $b$ is $C$ .

Keep the sides fixed and change only their angle again.

Matching opposite sides and angles

An opposite side–angle pair makes the sine rule available; an adjacent side–angle pair does not.

From the two area expressions $\tfrac12bc\sin A=\tfrac12ca\sin B$ , divide by $abc/2>0$ . This gives $\sin A/a=\sin B/b$ ; the remaining pair follows in the same way.

Sine rule

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

Use positive sides opposite their matching angles in a triangle with $A+B+C=180^\circ$ . Either reciprocal form is valid.

Not in the formula sheet: learn it.

With two angles and one side known, the triangle is fixed up to reflection and position. Carry unrounded calculator values and give the final accuracy requested; MEI otherwise expects accuracy appropriate to the context.

Worked example 3: calculate an opposite side

Question. A triangle has $A=65^\circ$ , $B=40^\circ$ and $a=9\,\mathrm{cm}$ . Find $b$ to $3$ significant figures.

Plan. Keep the complete pair $(9,65^\circ)$ together.

\begin{aligned} \frac{b}{\sin40^\circ}&=\frac9{\sin65^\circ} &&\text{match opposite pairs}\\ b&=\frac{9\sin40^\circ}{\sin65^\circ} &&\text{rearrange}\\ &=6.3831389\ldots &&\text{evaluate} \end{aligned}

Why does $\sin65^\circ$ divide the expression? It belongs to the known side, not the unknown side.

Answer. $b=6.38\,\mathrm{cm}$ .

Check. $B<A$ , so its opposite side is smaller than $9$ .

Change only $B$ , from $40^\circ$ to $55^\circ$ . The same known pair still sets the scale.

Worked example 4: retain the known pair

Question. Find $b$ to $3$ significant figures when $A=65^\circ$ , $B=55^\circ$ , $a=9\,\mathrm{cm}$ .

Plan. Replace only the sine associated with the unknown side.

\begin{aligned} b&=\frac{9\sin55^\circ}{\sin65^\circ} &&\text{sine rule}\\ &=8.1345085\ldots &&\text{evaluate without early rounding}\\ &\approx8.13 &&\text{round as requested} \end{aligned}

Why is the side larger than before? Its opposite angle increased, while the known opposite pair stayed fixed.

Answer. $b=8.13\,\mathrm{cm}$ .

Check. It is larger than $6.38$ and smaller than $9$ , in the same order as the opposite angles.

The rule also finds an angle, but inverse sine supplies only one candidate. Use the stated triangle information to decide whether a supplementary value fits.

Worked example 5: an admissible interior angle

Question. A triangle has $A=110^\circ$ , $a=12\,\mathrm{cm}$ , $b=8\,\mathrm{cm}$ . Find $B$ to $1$ decimal place.

Plan. Calculate $\sin B$ , then use the already obtuse angle to choose an admissible $B$ .

\begin{aligned} \sin B&=\frac{8\sin110^\circ}{12} &&\text{opposite pairs}\\ B&=38.789556\ldots^\circ &&\text{acute inverse-sine candidate} \end{aligned}

Its supplement is $141.210443\ldots^\circ$ , which cannot accompany $110^\circ$ : their sum exceeds $180^\circ$ .

Why check the angle sum? An inverse-trigonometric output is a candidate, not permission to ignore the rest of the geometry.

Answer. $B=38.8^\circ$ .

Check. The remaining angle is $31.210443\ldots^\circ>0$ , and the larger side $12$ faces the larger angle $110^\circ$ .

For another pairing $A=70^\circ$ , $B=45^\circ$ , $a=12$ gives $b=12\sin45^\circ/\sin70^\circ=9.0298478\ldots$ . The same method works; choose the rule from the information supplied, not from the orientation of the drawing.

Watch out

Wrong: pair an angle with the side beside it. Right: draw the side across the triangle from each angle and pair those two quantities.

Your matched side calculation changes the second angle again.

The cosine rule and a unique angle

Two sides and their included angle suggest the cosine rule; all three sides suggest its angle form.

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Mt3, Mt4 - Triangle area, sine rule and cosine rule

Area from two sides and their angle

Matching opposite sides and angles

The cosine rule and a unique angle

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