3.3 Applications of trigonometryIB Maths: Analysis and Approaches HL: Revision notes

Application questions describe the situation in words. Your first job is to turn the words into a labelled sketch on rough paper: mark every point, length and angle given, and show right angles where the ground is horizontal and a wall, tower or mast is vertical. Then pick triangles to solve one at a time. A typical chain is: find an angle using angles on a straight line or in a triangle, use the sine or cosine rule in a non-right-angled triangle, then finish in a right-angled triangle. Always state the triangle you are working in, for example 'in triangle $BFT$'.

If a triangle has a right angle, use Pythagoras's theorem $a^2+b^2=c^2$ and SOH CAH TOA. Many bearings problems contain a hidden right angle: sailing on $060^\circ$ then on $150^\circ$ is a turn of $90^\circ$, so the two legs are perpendicular and $HB=\sqrt{12^2+5^2}=13$ km. Give exact answers where the angle is $30^\circ$, $45^\circ$ or $60^\circ$ and no calculator is allowed: $\tan30^\circ=\frac{1}{\sqrt3}$, $\tan60^\circ=\sqrt3$, $\tan45^\circ=1$.

The angle of elevation is measured up from the horizontal to the line of sight. The angle of depression is measured down from the horizontal to the line of sight. Because the horizontal at the top and the ground are parallel, the angle of depression from $T$ to $Q$ equals the angle of elevation from $Q$ to $T$ (alternate angles). Two-observer problems: if $A$ and $B$ are in line with the foot of a tower, with elevations $35^\circ$ and $50^\circ$, then $A\hat{B}T=130^\circ$ and $A\hat{T}B=15^\circ$. Use the sine rule in triangle $ABT$, then right-angled trigonometry in triangle $BFT$ for the height.

A bearing is an angle measured clockwise from north, written with three figures: $040^\circ$, $110^\circ$, $263^\circ$.

• The back bearing (the bearing of $P$ from $Q$ when you know the bearing of $Q$ from $P$) is the bearing $\pm180^\circ$.
• North lines at different points are parallel, so use co-interior or alternate angles to move angles between them.
• To find the angle inside the triangle at a turning point, compare the back bearing with the new bearing: sailing $040^\circ$ then $110^\circ$ gives $P\hat{Q}R=220^\circ-110^\circ=110^\circ$. The final answer to a bearing question must be a bearing, not just an angle inside the triangle.

• Right angle: Pythagoras or SOH CAH TOA.
• Two sides and the angle between them: cosine rule for the third side.
• A side with its opposite angle: sine rule.
• Area of a region: $\frac12ab\sin C$.

When using the sine rule for an angle, check whether it should be acute or obtuse; if the triangle already has an obtuse angle, the others must be acute. Keep full calculator values between steps; round only the final answer (3 s.f. for lengths, 1 d.p. for angles).