Trigonometry · Trigonometry: Similar Triangles, Right Triangles and Acute Triangles

Multi-Step Problems: Two Right Triangles, Navigation and Surveying

2.3 2.2About 75 minutes

Learning goals

I can solve problems that need two right triangles, by finding a shared side first.
I can describe directions using compass bearings such as N 23° E.
I can solve navigation and surveying problems with the primary trigonometric ratios and the Pythagorean theorem.
I can set up and solve an equation when the shared side is unknown.

Words to know

compass bearing
A direction given as an angle turned from north or south toward east or west, for example N 23° E.

true bearing
A direction measured clockwise from north as a three-digit angle, for example 023°.

shared side
A side that belongs to two triangles in the same diagram; it links the two calculations.

surveying
Measuring land: distances, heights and angles, often using triangles.

When one triangle is not enough

Real problems often hide two right triangles in one diagram. A surveyor might measure two angles from one spot, or one angle from two spots. A ship's navigator turns and sails a second leg. The key is to find the side the two triangles share, because that side carries information from one triangle to the other.

A plan for multi-step problems

Draw a large, labelled sketch. Mark right angles.
Find the two triangles and the side they share.
Solve the triangle that has enough information (two pieces, including a side).
Use that answer (unrounded) in the second triangle.
Answer the question that was asked, with units, and check that it makes sense.

Worked example

Example 1: How long is the antenna?

An observer stands 50 m from the base of a building. The angle of elevation to the bottom of an antenna on the roof is 32°, and to the top of the antenna is 38°. How long is the antenna? (Ignore the observer's height.)

Two right triangles share the 50 m horizontal side.
Small triangle: building height y = 50 tan 32° = 31.24... m.
Large triangle: height to the top y + a = 50 tan 38° = 39.06... m.
Antenna: a = 50 tan 38° − 50 tan 32° = 7.82... m.

The antenna is about 7.8 m long.

Example 1, drawn to scale. Both right triangles share the 50 m horizontal side. The antenna length a is the difference of the two heights.

Worked example

Example 2: Two angles from two places

From point C on level ground, the angle of elevation to the top of a tower is 40°. From point D, 30 m farther away in a straight line, it is 25°. How tall is the tower?

Let h be the height and x the distance from C to the tower. The triangles share the side h, but we do not know it yet.
Triangle from C: tan 40° = h/x, so h = x tan 40°.
Triangle from D: tan 25° = h/(x + 30), so h = (x + 30) tan 25°.
Both equal h: x tan 40° = (x + 30) tan 25°. Expand: x tan 40° = x tan 25° + 30 tan 25°.
Collect x: x(tan 40° − tan 25°) = 30 tan 25°, so x = 30 tan 25° ÷ (tan 40° − tan 25°) = 37.53... m.
h = x tan 40° = 37.53... × 0.8391 = 31.49... m.

The tower is about 31.5 m tall.

Example 2, drawn to scale. Two overlapping right triangles share the height h. Neither triangle can be solved alone, so we write h two ways and set them equal.

Directions: compass bearings

Navigators describe directions with bearings. A compass bearing such as N 23° E means: face north, then turn 23° toward east. S 40° W means face south and turn 40° toward west. (A true bearing gives the angle clockwise from north as three digits, so N 23° E is the true bearing 023°.)

Worked example

Example 3: Sailing two legs

A sailboat leaves a dock and sails 12 km north, then 5 km east. How far is it from the dock, and what is its bearing from the dock?

North and east are perpendicular, so the path makes a right triangle with legs 12 km and 5 km.
Distance: √(12² + 5²) = √169 = 13 km.
Angle θ at the dock between north and the line to the boat: tan θ = 5/12, so θ ≈ 22.6° ≈ 23°.
The boat is 23° east of north from the dock.

The boat is 13 km from the dock, on a bearing of N 23° E.

Left: the two legs of the trip and the straight-line distance back to the start. Right: the bearing N 23° E means start facing north and turn 23° toward east.

Worked example

Example 4: Two boats from a lighthouse

From the top of a lighthouse 20 m above the water, the angles of depression to two boats directly in line with the lighthouse are 30° and 18°. How far apart are the boats?

Use the equal angles of elevation at the boats. The height 20 m is the shared side.
Near boat (larger angle, 30°): d₁ = 20 ÷ tan 30° = 34.64... m.
Far boat (18°): d₂ = 20 ÷ tan 18° = 61.55... m.
Distance apart: d₂ − d₁ = 26.91... m.

The boats are about 26.9 m apart.

Just for fun

Before satellites, navigators really did find their way across oceans using angles and triangles. Today your phone's GPS does the math for you, but if its battery dies on a canoe trip, trigonometry still works.

Hands-on activity

Treasure map with bearings

You need: a protractor, a ruler, plain or graph paper, a pencil, and a partner (optional: a compass app or a real compass outdoors).

Mark a START point on your paper and draw an arrow pointing to the top of the page for north.
Write two directions for a hidden "treasure", for example: walk 6 cm N 30° E, then 4 cm S 60° E. (These two directions are at right angles to each other. Check with your protractor!)
Swap papers with your partner, who draws the path with a protractor and ruler.
Both of you calculate the straight-line distance and bearing from START to the treasure, then measure them on the drawing.

What to look for: for the example path, the legs are perpendicular, so the distance is √(6² + 4²) ≈ 7.2 cm, and the angle at START between the first leg and the line to the treasure is tan⁻¹(4/6) ≈ 34°, so the bearing is about N 64° E (30° + 34°). Measured values within about 2 mm and 2° are good.

Your turn

An observer is 40 m from a building. The angle of elevation to the bottom of a flagpole on the roof is 41°, and to the top is 46°. How long is the flagpole?

From the top of a 35 m building, the angles of depression to two cars parked in a straight line from the building are 52° and 28°. How far apart are the cars?

A plane flies 300 km east, then 400 km south. How far is it from its start, and on what bearing from the start is it?

A hiker walks 3.2 km west, then 4.5 km north. Find her distance and bearing from the start.

From point C the angle of elevation to the top of a tower is 50°. From D, 30 m farther back in line, it is 35°. How tall is the tower?

A ship sails 25 km on a bearing of N 40° E. How far north and how far east of its start is it?

More practice: warm-up

Describe how to face the direction N 30° E.

A boat goes 6 km north and then 8 km east. How far is it from where it started?

From 50 m away, the angle of elevation to the roof of a building is 32°. How tall is the building (ignore eye height)?

In a two-triangle problem, what links the two triangles?

Describe the direction S 20° W.

More practice: on your own

From a point on level ground 60 m from a building, the angles of elevation to the bottom and top of a sign on the wall are 20° and 26°. How tall is the sign?

A ferry travels 18 km on a bearing of N 52° E. How far north and how far east does it go?

From the roof of a 20 m building, the angle of depression to the base of a taller building is 25°, and the angle of elevation to its top is 35°. How tall is the taller building?

A 7 m ladder reaches 6.5 m up a wall. Its foot slides out until the top is 6 m up. How far did the foot slide?

A lighthouse is 40 m tall. The angles of depression to two boats in line with it are 20° and 12°. How far apart are the boats?

Challenge

Two observers stand on opposite sides of a tall tower, 1000 m apart, in line with its base. The angles of elevation to the top are 40° and 55°. How tall is the tower?

A plane flies level at 2000 m directly toward a landmark. The angle of depression to the landmark changes from 10° to 25°. How far did the plane fly, to the nearest tenth of a kilometre?

A boat sails 20 km on a bearing of N 30° E, then turns and sails 15 km on a bearing of S 60° E. Show that the turn was 90°, then find the boat's distance and bearing from its start.

Check your understanding

Quiz

From 50 m away, the angles of elevation to the bottom and top of an antenna are 32° and 38°. How long is the antenna?

What does the bearing N 23° E mean?

A boat sails 12 km north and then 5 km east. How far is it from the start?

In a problem with two right triangles, what should you usually find first?

From a 20 m lighthouse, the angles of depression to two boats in line are 30° and 18°. How far apart are they?

A ship sails 25 km on a bearing of N 40° E. How far north of the start is it?

For parents and teachers

Big idea: Expectation 2.3: solving real-life problems in surveying and navigation with the primary trigonometric ratios and the Pythagorean theorem.

How to help: Insist on a big, clear sketch with the shared side highlighted. For bearings, stand up and act it out: face north, turn toward east by the given angle. A phone compass app can make this concrete.

Common difficulties:

Trying to solve the wrong triangle first (the one without enough information).
Subtracting instead of adding distances (or the reverse) when two triangles overlap.
Reading bearings backwards (N 23° E is not E 23° N).
Algebra in Example 2: collecting the x-terms. Review it slowly; it is the same kind of equation solving your teen practised in Grade 9.