3.3 Applications of trigonometryIB Maths: Applications and Interpretation HL: Revision notes

In a right-angled triangle, label the sides from the angle $\theta$ you are using: the hypotenuse (opposite the right angle), the opposite side and the adjacent side. $\sin\theta=\frac{\text{opp}}{\text{hyp}},\quad\cos\theta=\frac{\text{adj}}{\text{hyp}},\quad\tan\theta=\frac{\text{opp}}{\text{adj}},\qquad a^2+b^2=c^2.$ Use Pythagoras' theorem when two sides are known and you need the third, and $\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$ on your GDC for an angle. Set the GDC to degrees. Worked example: a ramp rises 1.2 m over a horizontal distance of 5 m. Its angle to the horizontal is $\tan^{-1}\left(\frac{1.2}{5}\right)=13.5^\circ$ and its length is $\sqrt{1.2^2+5^2}=5.14$ m.

The angle of elevation is the angle above the horizontal when you look up at an object. The angle of depression is the angle below the horizontal when you look down. Both are measured from a horizontal line, never from the vertical. The horizontal through the observer is parallel to the ground, so the angle of depression from the top of a cliff to a boat equals the angle of elevation from the boat to the top (alternate angles). Example: from a cliff 60 m high the angle of depression of a boat is $25^\circ$. The angle inside the triangle at the boat is also $25^\circ$, so the horizontal distance is $\frac{60}{\tan25^\circ}=129$ m.

A bearing is an angle measured clockwise from north, written with three digits, e.g. $070^\circ$ or $250^\circ$.

• To find the bearing back from $B$ to $A$, add or subtract $180^\circ$ (north lines are parallel, so co-interior angles sum to $180^\circ$).
• A journey of $d$ km on bearing $\theta$ gives an eastward distance $d\sin\theta$ and a northward distance $d\cos\theta$. Example: 15 km on bearing $070^\circ$ gives $15\sin70^\circ=14.1$ km east and $15\cos70^\circ=5.13$ km north. The bearing back is $070^\circ+180^\circ=250^\circ$.

For any triangle with sides $a,b,c$ opposite angles $A,B,C$: $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\quad\text{(sine rule)}$ $c^2=a^2+b^2-2ab\cos C,\quad\cos C=\frac{a^2+b^2-c^2}{2ab}\quad\text{(cosine rule)}$ $\text{Area}=\frac12ab\sin C.$ Use the sine rule when you know a side and its opposite angle plus one more side or angle. Use the cosine rule for two sides and the included angle, or for all three sides. In the area formula $C$ must be the angle between $a$ and $b$. Example: $AB=30$, $BC=40$ and $A\hat{B}C=110^\circ$ give $AC^2=900+1600-2400\cos110^\circ=3321$, so $AC=57.6$.

Most applications are given in words, so the first step is a labelled diagram.

1. Draw a rough sketch showing the points, in the order of the statement.
2. Mark a north line at each point for bearings, and a horizontal line for elevation or depression.
3. Label every given length and angle, and mark the unknown with a letter.
4. Find the triangle (right-angled or not) that contains the unknown, and add any angle you can deduce (angle sum $180^\circ$, alternate angles, back bearing). Several triangles may be linked: solve one, then use its answer in the next. Keep full GDC values between steps and round only the final answer to 3 significant figures (angles to 1 decimal place).