Applications

Measuring the Unreachable

Trigonometry earns its keep when you cannot measure something directly: the height of a building, the width of a river, the distance to a ship at sea.

The whole skill is this: turn the situation into a triangle, then use the tool that fits.

Angle of Elevation and Depression

Most real problems start with a line of sight and the angle it makes with the horizontal.

  • Angle of elevation: how far you look up from horizontal
  • Angle of depression: how far you look down from horizontal

Notice what the animation shows: as the observer walks closer, the distance shrinks and the angle grows, yet dtan⁡θd\tan\theta always returns the same height. The triangle changes, but the building does not.

The Modeling Steps

Every one of these problems follows the same recipe:

  1. Draw the scene as a triangle
  2. Label what you know: which sides, which angles
  3. Choose the tool
  4. Solve for the missing piece

Right triangle? Use SOH-CAH-TOA. No right angle? Use the law of sines or law of cosines.

Example: Height by Elevation

You stand 3030 m from the base of a tree and look up to its top at an angle of elevation of 40°40°. How tall is the tree?

The distance and height form a right triangle. Height is opposite, distance is adjacent, so use tangent:

tan⁡40°=h30h=30tan⁡40°≈25.2 m\begin{aligned} \tan 40° &= \frac{h}{30} \\[0.5em] h &= 30\tan 40° \\[0.5em] &\approx 25.2 \text{ m} \end{aligned}

Example: Distance with the Law of Cosines

A ship sails 2020 km, turns 50°50° from its course, then sails another 1515 km. How far is it from the start?

The turn creates a triangle. The interior angle is 180°−50°=130°180° - 50° = 130°. Two sides and the angle between them is an SAS setup, so use the law of cosines:

c2=202+152−2(20)(15)cos⁡130°=625−600(−0.643)≈1010.6\begin{aligned} c^2 &= 20^2 + 15^2 - 2(20)(15)\cos 130° \\[0.5em] &= 625 - 600(-0.643) \\[0.5em] &\approx 1010.6 \end{aligned}

So c≈31.8c \approx 31.8 km from the start.

Example: Distance with the Law of Sines

Two fire lookouts sit 1010 km apart. Each measures the angle to a distant fire: 65°65° from one, 50°50° from the other. How far is the fire from the first lookout?

The third angle is 180°−65°−50°=65°180° - 65° - 50° = 65°. The 1010 km side is opposite it, giving a complete pair (AAS), so use the law of sines:

xsin⁡50°=10sin⁡65°x=10sin⁡50°sin⁡65°≈8.5 km\begin{aligned} \frac{x}{\sin 50°} &= \frac{10}{\sin 65°} \\[0.5em] x &= \frac{10\sin 50°}{\sin 65°} \\[0.5em] &\approx 8.5 \text{ km} \end{aligned}

A Word on Bearings

Navigation problems often give directions as bearings: an angle measured clockwise from north.

  • Due north is 000°000°
  • Due east is 090°090°
  • Due south is 180°180°

Turn the bearings into the interior angles of a triangle, and the same two laws solve the rest.