Trigonometric Ratios

Worked example 1 — the same triangle, two viewpoints

Triangle $ABC$ is right-angled at $B$, with $AB=12\text{ cm}$, $BC=5\text{ cm}$ and $AC=13\text{ cm}$. Write the three trigonometric ratios for angle $A$, then find angle $A$.

For angle $A$, $BC$ is opposite and $AB$ is adjacent. Therefore:

$$\sin A=\frac5{13},\qquad \cos A=\frac{12}{13},\qquad \tan A=\frac5{12}.$$

Use an inverse trigonometric function to recover the angle: $A=\sin^{-1}(5/13)=22.6198\ldots^\circ$, so $A=22.6^\circ$ to 1 decimal place.

Check: $\cos^{-1}(12/13)$ and $\tan^{-1}(5/12)$ give the same angle. For angle $C$, the opposite side is instead 12 cm and $\sin C=12/13$. Also $A+C=90^\circ$.

Worked example 2 — vertical and horizontal components

A straight 8 m support makes an angle of $35^\circ$ with horizontal ground. Find the vertical height $h$ of its top and the horizontal distance $d$ from its bottom to the point directly beneath its top. Give lengths to 3 significant figures.

The support is the hypotenuse. Relative to $35^\circ$, the height is opposite and the ground distance is adjacent. Use sine for $h$ and cosine for $d$:

$$\sin35^\circ=\frac h8\quad\Rightarrow\quad h=8\sin35^\circ=4.588611\ldots\text{ m}\approx4.59\text{ m}.$$

$$\cos35^\circ=\frac d8\quad\Rightarrow\quad d=8\cos35^\circ=6.553216\ldots\text{ m}\approx6.55\text{ m}.$$

Multiply by 8 because it is the denominator in each equation. If instead the unknown hypotenuse were in the denominator, as in $\sin35^\circ=4/x$, rearrange to $x=4/\sin35^\circ$.

Check: both sides are shorter than 8 m. Using the unrounded values, $h^2+d^2=64=8^2$. The height is shorter than the ground distance because $35^\circ<45^\circ$.

Worked example 3 — choose tangent when no hypotenuse is given

A right-angled triangle has opposite side 9 cm and adjacent side 12 cm relative to acute angle $\theta$. Find $\theta$ to 1 decimal place.

The two supplied sides match tangent directly. There is no need to find the hypotenuse first:

$$\tan\theta=\frac9{12}=0.75\quad\Rightarrow\quad \theta=\tan^{-1}(0.75)=36.8698\ldots^\circ\approx36.9^\circ.$$

The calculator’s $\tan^{-1}$ key asks “which acute angle has tangent 0.75?” It does not mean $1/\tan(0.75)$.

Check: the opposite side is smaller than the adjacent side, so the angle should be below $45^\circ$. Substituting the unrounded angle gives $\tan\theta=0.75$. The rounded angle gives a close approximation, not an exact equality.

Worked example 4 — two positions on level ground

A vertical mast has base $B$ and top $T$. Points $P$ and $Q$ lie on the same straight level line from $B$, with $BP=18\text{ m}$, $BQ=30\text{ m}$, and $P$ between $B$ and $Q$. The angle between $PT$ and the ground is $32^\circ$. Find the mast’s height and the angle $TQB$.

1. In $PBT$, use tangent: $BT=18\tan32^\circ=11.247648\ldots\text{ m}$. The mast’s height is $11.2\text{ m}$ to 3 significant figures.
2. In $QBT$, the adjacent side is the whole 30 m, not $PQ=12\text{ m}$. Thus $\tan\angle TQB=BT/30$.
3. Keep the unrounded height: $\angle TQB=\tan^{-1}((18\tan32^\circ)/30)=20.5521\ldots^\circ\approx20.6^\circ$ to 1 decimal place.

Check: moving farther from the same mast gives a smaller angle, and $20.6^\circ<32^\circ$. The problem measures the angle from ground level; if an observer’s eye level were given, we would first find the height above that eye level.