1.05b, 1.05c - Triangle area, sine rule and cosine rule | OCR A-Level Mathematics A

Choose and use trigonometric rules for triangles that need not contain a right angle, check when the same data allow two triangles, and translate bearings into the angles a calculation needs.

Area from two sides and their angle

Two rigid rods of lengths $8\,\mathrm{cm}$ and $11\,\mathrm{cm}$ form two sides of a triangle. Must opening their included angle from $60^\circ$ to $120^\circ$ increase the area? Surprisingly, these two triangles have exactly the same area. Their heights, rather than the apparent width of the opening, explain why.

You already know

A triangle's area is half its base times its perpendicular height. In a right triangle, a length $L$ at angle $\theta$ to the horizontal has height $L\sin\theta$ ; for example, $5\sin30^\circ=2.5$ .

Use lower-case letters for sides opposite the matching capital-letter angles. Thus side $a$ faces $A$ . Sides $a$ and $b$ meet at $C$ , so taking $b$ as base gives height $a\sin C$ .

Triangle area

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

The angle must be included between the two selected positive side lengths. For a triangle, $0^\circ<C<180^\circ$ .

Not in the formula sheet: learn it.

For an obtuse angle the perpendicular height falls outside the triangle, but $\sin(180^\circ-C)=\sin C$ , so its length and the area remain positive.

Worked example 1: an acute included angle

Question. Find the exact area when sides $8\,\mathrm{cm}$ and $11\,\mathrm{cm}$ meet at $60^\circ$ .

Plan. Use the two given sides with the angle between them.

\begin{aligned} \Delta&=\frac12(8)(11)\sin60^\circ &&\text{included-angle formula}\\ &=44\left(\frac{\sqrt{3}}2\right) &&\text{exact sine}\\ &=22\sqrt{3} &&\text{simplify} \end{aligned}

Why use $60^\circ$ ? It lies between the two measured sides; an angle opposite either side would not fit this substitution.

Answer. $22\sqrt{3}\,\mathrm{cm}^2$ , approximately $38.1\,\mathrm{cm}^2$ to $3$ significant figures.

Check. The height on base $11$ is $8\sin60^\circ=4\sqrt{3}$ , giving the same area.

Change only the included angle to its supplement. The height remains unchanged, even though the third side does not.

Worked example 2: an obtuse included angle

Question. Find the exact area for the same sides meeting at $120^\circ$ .

Plan. Use the obtuse angle directly; its sine is positive.

\begin{aligned} \Delta&=\frac12(8)(11)\sin120^\circ &&\text{same formula}\\ &=44\left(\frac{\sqrt{3}}2\right) &&\sin120^\circ=\sin60^\circ\\ &=22\sqrt{3} &&\text{simplify} \end{aligned}

Why is there no negative area? A triangle's height is a positive length; the sine of an obtuse interior angle is positive too.

Answer. Again $22\sqrt{3}\,\mathrm{cm}^2$ , or $38.1\,\mathrm{cm}^2$ to $3$ significant figures.

Check. Both answers are below $44\,\mathrm{cm}^2$ , the maximum for these sides, attained at $90^\circ$ .

The opening puzzle is resolved: the two angles give equal heights. The largest area occurs between them, at a right angle.

Watch out

Wrong: $\Delta=\tfrac12ab\sin A$ when $A$ is opposite side $a$ . Right: the angle between $a$ and $b$ is $C$ . Label the opposite pairs first.

Change the included angle once more and keep the answer exact.

Matching opposite sides and angles

The sine rule is useful when you know at least one complete side–opposite-angle pair.

Equate the area forms $\tfrac12bc\sin A=\tfrac12ca\sin B$ . Dividing by the positive product $abc/2$ gives $\sin A/a=\sin B/b$ ; the third pair follows in the same way.

Sine rule

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

Sides are positive and opposite their matching angles; $A+B+C=180^\circ$ . The reciprocal form is equally valid.

Not in the formula sheet: learn it.

With two angles known, the third is fixed, so there is one triangle up to reflection and movement. Finding a side then needs just one rearrangement.

Worked example 3: find a side

Question. In a triangle, $A=70^\circ$ , $B=45^\circ$ and $a=12\,\mathrm{cm}$ . Find $b$ to $3$ significant figures.

Plan. Pair $b$ with $45^\circ$ and $12$ with $70^\circ$ .

\begin{aligned} \frac{b}{\sin45^\circ}&=\frac{12}{\sin70^\circ} &&\text{opposite pairs}\\ b&=\frac{12\sin45^\circ}{\sin70^\circ} &&\text{isolate }b\\ &=9.0298478\ldots &&\text{evaluate without early rounding} \end{aligned}

Why put $\sin70^\circ$ underneath? The known side is opposite $70^\circ$ ; retaining that pair prevents the common inverted ratio.

Answer. $b=9.03\,\mathrm{cm}$ .

Check. $45^\circ<70^\circ$ , so $b<12$ , as expected.

Change only $B$ , from $45^\circ$ to $60^\circ$ . Its opposite side should grow, while still remaining shorter than $a$ .

Worked example 4: a larger opposite angle

Question. Find $b$ when $A=70^\circ$ , $B=60^\circ$ and $a=12\,\mathrm{cm}$ , to $3$ significant figures.

Plan. Keep the known pair and replace only the numerator's sine.

\begin{aligned} b&=\frac{12\sin60^\circ}{\sin70^\circ} &&\text{sine-rule pairing}\\ &=11.0592598\ldots &&\text{evaluate}\\ &\approx11.1 &&\text{round once} \end{aligned}

Why does the denominator stay fixed? The known side and its opposite angle have not changed.

Answer. $b=11.1\,\mathrm{cm}$ .

Check. $9.03<11.1<12$ , matching the ordering of $45^\circ,60^\circ,70^\circ$ .

When instead finding an angle from a sine value, you must examine the supplementary candidate too; the ambiguity part teaches that decision.

Exam tip

Write the paired substitution before giving a decimal. It shows the sine-rule method even if the final arithmetic slips. Keep guard digits until the requested final precision.

Now change the second angle again.

The cosine rule and a unique angle

Use the cosine rule for two sides and their included angle, or for three sides, when no complete opposite pair is available.

Put the included angle $C$ at the origin. The endpoints of the adjacent sides are $(b,0)$ and $(a\cos C,a\sin C)$ . The squared distance between them is

\begin{aligned} c^2&=(b-a\cos C)^2+(a\sin C)^2\\ &=a^2+b^2-2ab\cos C \end{aligned}

using $\cos^2C+\sin^2C=1$ from the unit circle. At $C=90^\circ$ , the cosine term vanishes and Pythagoras returns.

Cosine rule

c^2=a^2+b^2-2ab\cos C,\qquad \cos C=\frac{a^2+b^2-c^2}{2ab}.

Use positive sides in a nondegenerate triangle and $0^\circ<C<180^\circ$ . Each side must be smaller than the sum of the other two.

Not in the formula sheet: learn it.

The subtracted cosine term can be positive or negative. An obtuse included angle makes the opposite side longer, not shorter.

Worked example 5: the third side

Question. Find the exact third side for $a=7$ , $b=9$ and $C=60^\circ$ .

Plan. Find the squared side first, then take the positive root.

\begin{aligned} c^2&=7^2+9^2-2(7)(9)\cos60^\circ &&\text{cosine rule}\\ &=130-63=67 &&\text{use exact cosine}\\ c&=\sqrt{67} &&\text{positive length} \end{aligned}

Why take only the positive root? A side length cannot be negative; the algebraic negative root is not another triangle.

Answer. $c=\sqrt{67}$ , approximately $8.19$ .

Check. $2<\sqrt{67}<16$ , satisfying the triangle inequalities.

Change only the included angle to $120^\circ$ . Its cosine changes sign, so the same correction term now increases the squared length.

Worked example 6: the obtuse comparison

Question. Find the exact third side for $a=7$ , $b=9$ and $C=120^\circ$ .

Plan. Retain the negative sign in the rule and substitute the negative cosine carefully.

\begin{aligned} c^2&=130-126\left(-\frac12\right) &&\text{exact obtuse cosine}\\ &=193 &&\text{subtract a negative}\\ c&=\sqrt{193} &&\text{positive root} \end{aligned}

Why is this side longer? Opening the angle separates the side endpoints; the cosine rule encodes that geometry.

Answer. $c=\sqrt{193}$ , approximately $13.9$ .

Check. It exceeds both adjacent sides and remains below their sum $16$ .

For three-side data, rearrange the same rule. Cosine decreases from $1$ to $-1$ across the possible interior angles, so it gives a unique angle.

Worked example 7: recover an angle

Question. Find angle $C$ for sides $a=7$ , $b=9$ , $c=10$ , to $3$ significant figures.

Plan. Use all three sides to calculate cosine, then take the angle in $(0^\circ,180^\circ)$ .

\begin{aligned} \cos C&=\frac{49+81-100}{126}=\frac5{21} &&\text{rearranged rule}\\ C&=\cos^{-1}\left(\frac5{21}\right)=76.225853\ldots^\circ &&\text{interior angle} \end{aligned}

Why not use the sine rule? No side has a known opposite angle yet.

Answer. $C=76.2^\circ$ .

Check. Since $10^2<7^2+9^2$ , the angle opposite $10$ must be acute, as found.

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1.05b, 1.05c - 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

Continue this lesson.

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