4.8 - Right-angled trigonometry and bearings

4.8 - Right-angled trigonometry and bearings

You will connect an acute angle to the side ratios of a right triangle, calculate missing lengths and angles, and use these methods with bearings. Bearings turn a journey into perpendicular north–south and east–west distances. Keep your calculator in degree mode throughout.

Sine, cosine and tangent

A boat travels 12 km12\,\mathrm{km} on bearing 030∘030^\circ, then 5 km5\,\mathrm{km} due east. How far is it from its start, and what bearing points straight to it? You can answer without sailing the route: start by understanding the triangle inside the first leg.

You already know

The hypotenuse is opposite a right angle. Similar triangles have equal corresponding angles and equal ratios of corresponding lengths. The two acute angles of a right triangle total 90∘90^\circ.

For a 5,12,135,12,13 right triangle, 5/135/13 compares one shorter side with the hypotenuse. Enlarge every length and the ratio stays unchanged. This is why a ratio can describe an angle without specifying the triangle's size.

Opposite and adjacent

Relative to a chosen acute angle θ\theta, the opposite side is across from θ\theta. The adjacent side touches θ\theta and is not the hypotenuse. Choosing the other acute angle swaps opposite and adjacent; the hypotenuse stays the same.

Name the sides before selecting a ratio. A sloping side is not automatically the hypotenuse: the right angle decides that.

[DIAGRAM: asset_slug: m070_ratio_triangle; caption: Opposite adjacent and hypotenuse relative to the chosen acute angle theta; description: A right triangle marks theta at its left vertex and the right angle at the bottom right. Opposite is the vertical leg and adjacent the horizontal leg. The schematic cannot supply measurements.; file: m070_ratio_triangle.png; recommended_method: matplotlib]
Opposite adjacent and hypotenuse relative to the chosen acute angle theta

Sine, cosine and tangent are ratios of lengths in the same unit, so their units cancel.

Right-triangle ratios

For an acute angle θ\theta in a right triangle,

sin⁡θ=oppositehypotenuse,cos⁡θ=adjacenthypotenuse,tan⁡θ=oppositeadjacent\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}},\qquad \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}},\qquad \tan\theta=\frac{\text{opposite}}{\text{adjacent}}

SOH CAH TOA is a useful reminder. Absent from both the Foundation and Higher sheets. Not in the formulae sheet: learn it.

For an acute angle, all three ratios are positive. Sine and cosine are less than 11, while tangent can exceed 11.

Worked example 1: use the side names

Question. Relative to θ\theta, a right triangle has opposite 5 cm5\,\mathrm{cm}, adjacent 12 cm12\,\mathrm{cm} and hypotenuse 13 cm13\,\mathrm{cm}. Write all three trigonometric ratios.

Plan. Put the named sides into SOH CAH TOA.

sin⁡θ=513opposite over hypotenusecos⁡θ=1213adjacent over hypotenusetan⁡θ=512opposite over adjacent\begin{aligned} \sin\theta&=\frac{5}{13} &&\text{opposite over hypotenuse}\\ \cos\theta&=\frac{12}{13} &&\text{adjacent over hypotenuse}\\ \tan\theta&=\frac{5}{12} &&\text{opposite over adjacent} \end{aligned}

Why is tangent's denominator 1212? Tangent uses the adjacent side, not the hypotenuse.

Answer. sin⁡θ=5/13\sin\theta=5/13, cos⁡θ=12/13\cos\theta=12/13, tan⁡θ=5/12\tan\theta=5/12.

Check. 52+122=1325^2+12^2=13^2, so these lengths form a right triangle.

Change only the chosen angle. Let ϕ\phi be the other acute angle of this same triangle. The side lengths stay fixed, but the opposite and adjacent roles swap.

Worked example 2: look from the other angle

Question. For the same triangle, write all three ratios relative to ϕ=90∘−θ\phi=90^\circ-\theta.

Plan. Swap the shorter-side roles while keeping the hypotenuse.

sin⁡ϕ=1213new opposite over hypotenusecos⁡ϕ=513new adjacent over hypotenusetan⁡ϕ=125new opposite over adjacent\begin{aligned} \sin\phi&=\frac{12}{13} &&\text{new opposite over hypotenuse}\\ \cos\phi&=\frac{5}{13} &&\text{new adjacent over hypotenuse}\\ \tan\phi&=\frac{12}{5} &&\text{new opposite over adjacent} \end{aligned}

Why is 1313 unchanged? It remains opposite the right angle, whichever acute angle you choose.

Answer. sin⁡ϕ=12/13\sin\phi=12/13, cos⁡ϕ=5/13\cos\phi=5/13, tan⁡ϕ=12/5\tan\phi=12/5.

Check. tan⁡ϕ>1\tan\phi>1, consistent with its opposite side being longer than its adjacent side.

A useful surprise is visible here: the sine of one acute angle equals the cosine of the other. Both compare the same shorter side with the same hypotenuse. Thus sin⁡θ=cos⁡(90∘−θ)\sin\theta=\cos(90^\circ-\theta), for example sin⁡30∘=cos⁡60∘=1/2\sin30^\circ=\cos60^\circ=1/2.

Watch out

The line tan⁡θ=5/13\tan\theta=5/13 uses the hypotenuse. That is sine. The correct tangent ratio is 5/125/12.

Try a new triangle; the question supplies the side roles directly.

Calculating missing lengths

Choose the ratio containing the known length and the wanted length. Write an equation before using the calculator. The equation tells you whether to multiply or divide.

For instance, sin⁡θ=o/h\sin\theta=o/h gives o=hsin⁡θo=h\sin\theta when the hypotenuse is known, but h=o/sin⁡θh=o/\sin\theta when the opposite is known. Similarly, tan⁡θ=o/a\tan\theta=o/a gives o=atan⁡θo=a\tan\theta or a=o/tan⁡θa=o/\tan\theta. These rearrangements are not supplied on either tier's sheet.

Worked example 3: the unknown denominator

Question. A right triangle has angle 38∘38^\circ and adjacent side 10 cm10\,\mathrm{cm}. Find the hypotenuse to 33 significant figures.

Plan. Use cosine because the two relevant sides are adjacent and hypotenuse.

cos⁡38∘=10hadjacent over hypotenusehcos⁡38∘=10multiply by hh=10cos⁡38∘divide by cosine=12.690182… cmdegree mode\begin{aligned} \cos38^\circ&=\frac{10}{h} &&\text{adjacent over hypotenuse}\\ h\cos38^\circ&=10 &&\text{multiply by }h\\ h&=\frac{10}{\cos38^\circ} &&\text{divide by cosine}\\ &=12.690182\ldots\,\mathrm{cm} &&\text{degree mode} \end{aligned}

Why divide? The unknown is the denominator of the original ratio.

Answer. 12.7 cm12.7\,\mathrm{cm}.

Check. The hypotenuse exceeds 10 cm10\,\mathrm{cm}; its unrounded value times cos⁡38∘\cos38^\circ returns 1010.

Change only the role of the known 10 cm10\,\mathrm{cm} length: it is now the hypotenuse. Keep the angle 38∘38^\circ and the same two side roles.

Worked example 4: the unknown numerator

Question. A right triangle has angle 38∘38^\circ and hypotenuse 10 cm10\,\mathrm{cm}. Find the adjacent side to 33 significant figures.

Plan. Write the cosine ratio, now with the unknown in its numerator.

cos⁡38∘=a10adjacent over hypotenusea=10cos⁡38∘multiply by 10=7.880107… cmevaluate in degrees\begin{aligned} \cos38^\circ&=\frac{a}{10} &&\text{adjacent over hypotenuse}\\ a&=10\cos38^\circ &&\text{multiply by }10\\ &=7.880107\ldots\,\mathrm{cm} &&\text{evaluate in degrees} \end{aligned}

Why multiply this time? The known denominator is 1010; multiplying clears it.

Answer. 7.88 cm7.88\,\mathrm{cm}.

Check. 7.880107…/10=cos⁡38∘7.880107\ldots/10=\cos38^\circ, and the shorter side is below 10 cm10\,\mathrm{cm}.

Use the actual side roles rather than memorising “trigonometry means multiply”.

Watch out

In the first example, h=10cos⁡38∘h=10\cos38^\circ would make the hypotenuse shorter than a leg. Rearrange cos⁡38∘=10/h\cos38^\circ=10/h to obtain h=10/cos⁡38∘h=10/\cos38^\circ.

The next problem changes the known angle and length, and asks for the other shorter side.

Calculating acute angles

Two known sides let you calculate a ratio first, then recover the angle with an inverse function. Use sin⁡−1\sin^{-1}, cos⁡−1\cos^{-1} or tan⁡−1\tan^{-1} in degree mode. Here the superscript −1-1 means inverse function; it does not mean a reciprocal.

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