Applications of right-angled trigonometryIB MYP Maths Standard: Revision notes

In a right-angled triangle, label the sides from the angle $\theta$ you are using: hypotenuse (opposite the right angle), opposite and adjacent. $\sin\theta=\frac{\text{opp}}{\text{hyp}},\quad\cos\theta=\frac{\text{adj}}{\text{hyp}},\quad\tan\theta=\frac{\text{opp}}{\text{adj}}$ Use Pythagoras, $a^2+b^2=c^2$, when you know two sides and want the third. Use trigonometry when an angle is involved. To find an angle use $\sin^{-1}$, $\cos^{-1}$ or $\tan^{-1}$. Keep full calculator values and round only at the end.

Many problems need two or more right-angled triangles. Draw a clear sketch, mark known lengths, and decide which length to find first. Use the answer from the first triangle in the second one (keep the full calculator value). Example: a ship sails $40$ km on a bearing of $035^\circ$. North $=40\cos35^\circ=32.8$ km, east $=40\sin35^\circ=22.9$ km. If a lighthouse is $50$ km due north of the start, it is $50-32.8=17.2$ km north of the ship, so the distance to it is $\sqrt{22.9^2+17.2^2}\approx28.7$ km.

The angle of elevation is the angle above the horizontal, looking up at an object. The angle of depression is the angle below the horizontal, looking down. They are equal (alternate angles) when measured between the same two points. Example: a cable rises $350$ m over a horizontal distance of $1200$ m. The angle of elevation is $\tan^{-1}\frac{350}{1200}=16.3^\circ$ and the cable length is $\sqrt{1200^2+350^2}=1250$ m.

To find an angle or length in 3D, find a right-angled triangle and draw it separately in 2D. In a cuboid with base $8$ cm by $6$ cm and height $5$ cm:

• Base diagonal $AC=\sqrt{8^2+6^2}=10$ cm.
• Space diagonal $AG=\sqrt{10^2+5^2}=\sqrt{125}=11.2$ cm.
• The angle between a line and a plane is the angle between the line and its projection (shadow) on the plane. For $AG$ and the base, the projection is $AC$, so the angle is $G\hat{A}C=\tan^{-1}\frac{5}{10}=26.6^\circ$.

In a square-based pyramid with apex $V$ above the centre $M$ of the base, $VM$ is perpendicular to the base, so triangle $VMP$ is right-angled at $M$. Base side $10$ cm, height $12$ cm: $PM=\frac12\sqrt{10^2+10^2}=5\sqrt2=7.07$ cm. Then the sloping edge $VP=\sqrt{12^2+7.07^2}=13.9$ cm, and the angle between $VP$ and the base is $\tan^{-1}\frac{12}{7.07}=59.5^\circ$. Note: $PM$ is half the diagonal, not half the side.

Real situations are modelled with straight lines and right-angled triangles. Always check your answer against the real-life rule, and say how accurate the model is. Example: if a cable can rise at most $20^\circ$ over $1200$ m, the greatest height is $1200\tan20^\circ=436.8$ m. A height of $437$ m would break the rule, so the whole-number limit is $436$ m. Always round in the direction that keeps the rule. Mention assumptions: a real cable sags, boats are pushed by currents and the wind changes the path of a plane.