C. Trigonometric Functions · Trigonometry of Acute Triangles

Real-World Problems: Surveying, Navigation and Construction

C1.5 C1.4About 75 minutes

Learning goals

I can turn a real-world situation into a labelled acute triangle.
I can solve surveying, navigation and building problems with the sine law and the cosine law.
I can use bearings to find an interior angle of a triangle.
I can answer the question that was asked, with units and sensible rounding.

Words to know

back bearing
The bearing for going the opposite way along a line: add or subtract 180°.

triangulation
Finding the position of a point by measuring angles to it from two known points.

truss
A rigid frame made of triangles, used to hold up roofs and bridges.

Triangles at work

Forest-fire lookouts, ferry captains, surveyors and carpenters all solve acute triangles. Expectation C1.5 is about these problems. The maths is what you already know; the new skill is setting up the triangle:

Read the problem twice. Draw a large sketch and label every given value.
Decide which values are sides and which are angles. Find any interior angles you need (from bearings, angle sums or parallel lines).
Classify the case (AAS, ASA, SSA, SAS, SSS) and choose the law.
Solve, then answer the question in a sentence with units.

Worked example

Example 1: Triangulating a forest fire

Two fire lookout towers, A and B, are 15 km apart. A ranger in tower A sees smoke at an angle of 52° from the line joining the towers; a ranger in B sees it at 61° from the same line. How far is the fire from each tower, and how far is it from the straight road that joins the towers?

Sketch △ABF. Known: ∠A = 52°, ∠B = 61° and the side between them AB = 15 km (ASA).
∠F = 180° − 52° − 61° = 67°. AB is across from ∠F: complete pair.
AF is across from ∠B: AF = 15 sin 61° ÷ sin 67° ≈ 14.3 km.
BF is across from ∠A: BF = 15 sin 52° ÷ sin 67° ≈ 12.8 km.
Distance from the road (perpendicular d): in the right triangle at A, d = AF sin 52° ≈ 11.2 km.

The fire is about 14.3 km from tower A, 12.8 km from tower B and 11.2 km from the road.

Example 1, to scale. Two angles measured from known points fix the fire's position: this is triangulation.

Worked example

Example 2: A two-leg boat trip

A boat leaves a harbour and sails 18 km on a bearing of 040°. It then turns and sails 12 km on a bearing of 150° to a buoy. How far is the buoy from the harbour, and on what bearing?

At the turning point, the direction back to the harbour is the back bearing 040° + 180° = 220°. The boat heads off at 150°. The interior angle between them is 220° − 150° = 70°.
Known: two sides (18 and 12) and the angle between them (70°): SAS, cosine law.
d² = 18² + 12² − 2(18)(12) cos 70° ≈ 468 − 147.75 ≈ 320.25, so d ≈ 17.9 km.
Angle at the harbour (across from the 12 km side): sin θ = 12 sin 70° ÷ 17.895 ≈ 0.6301, θ ≈ 39.1°.
The buoy is clockwise from the first leg, so the bearing is 040° + 39.1° ≈ 79°.

The buoy is about 17.9 km from the harbour on a bearing of about 079°.

Example 2. North lines at both points. The interior angle at the turn comes from the back bearing (220°) and the new heading (150°).

Worked example

Example 3: Pitch of a roof truss

A truss for a garden shed has a 7.0 m bottom chord and two rafters, 6.2 m and 5.4 m long, meeting at the peak. Find the angle each rafter makes with the bottom chord, and the peak angle.

SSS: cosine law.
Angle at the end of the 6.2 m rafter (across from 5.4): cos θ = (6.2² + 7.0² − 5.4²) ÷ (2 × 6.2 × 7.0) = 58.28 ÷ 86.8 ≈ 0.6714, θ ≈ 47.8°.
Angle at the end of the 5.4 m rafter (across from 6.2): cos φ = (5.4² + 7.0² − 6.2²) ÷ (2 × 5.4 × 7.0) = 39.72 ÷ 75.6 ≈ 0.5254, φ ≈ 58.3°.
Peak angle = 180° − 47.8° − 58.3° ≈ 73.9°.

The 6.2 m rafter makes about 48° with the chord, the 5.4 m rafter about 58°, and the peak angle is about 74°.

Example 3, to scale. The longer rafter is less steep; the shorter rafter rises at a steeper angle to reach the same peak.

Worked example

Example 4: Planning a tunnel

Engineers want a straight tunnel through a hill from A to B. From a point C where both ends can be seen, CA = 1.8 km, CB = 2.3 km and ∠ACB = 68°. How long will the tunnel be, and what angle should the tunnel make with line AC at A?

SAS: cosine law for AB.
AB² = 1.8² + 2.3² − 2(1.8)(2.3) cos 68° ≈ 8.53 − 3.102 ≈ 5.428, so AB ≈ 2.33 km.
Angle at A (across from CB = 2.3): sin A = 2.3 sin 68° ÷ 2.330 ≈ 0.9153, ∠A ≈ 66.2°.
Check the triangle is acute: ∠B ≈ 180° − 68° − 66.2° ≈ 45.8°. ✓

The tunnel is about 2.33 km long, and at A it should make an angle of about 66° with line AC.

Just for fun

A funny-but-true fact: before satellites, much of the mapping of large countries, including Canada, was done by triangulation: chains of triangles, with angles measured from hilltops and towers. Mapping a country one triangle at a time takes a lot of patience and a lot of sine laws.

Hands-on activity

Triangulate a spot in your yard or park

You need: a measuring tape, a protractor glued or taped to a piece of cardboard (or a phone compass app), two sticks or markers, a small target (a ball or a backpack) and a helper.

Push two sticks into the ground 5 m to 10 m apart. This is your baseline AB. Measure it.
Place the target somewhere off to one side.
Stand at A. Line up the protractor with the baseline and measure the angle between the baseline and the target. Repeat at B.
Use the sine law to calculate the distance from A to the target.
Measure it with the tape.

What to look for: your calculation within about 5% of the measured distance. Angles that are very small or close to 90° make the result less reliable. Write one sentence explaining why rangers want their towers far apart.

Your turn

Two lighthouses are 9.0 km apart. A ship is seen at 48° from the line between them at the first lighthouse and at 63° at the second. How far is the ship from the first lighthouse?

A plane flies 250 km on a bearing of 060°, then 180 km on a bearing of 170°. Find the interior angle at the turn and the plane's distance from its start.

A triangular lot has sides 48 m, 55 m and 61 m. Find the largest angle of the lot.

A crane's boom is 32 m long and its support cable is attached 20 m along the base from the boom's pivot. The angle at the pivot between the base and the boom is 62°. How long is the cable from the base point to the top of the boom?

A surveyor measures a lake: from point P, distances 640 m and 710 m to the ends of the lake and an angle of 54° between them. How long is the lake?

Two hikers leave camp. One walks 4.2 km on a bearing of 030°; the other walks 5.0 km on a bearing of 100°. How far apart are they?

More practice: warm-up

What is the back bearing of 065°?

What is the back bearing of 230°?

Two angles of a triangle are 55° and 72°. Find the third.

In ASA, which law do you use after the angle sum?

Name the method of locating a point by measuring angles from two known points.

More practice: on your own

A ferry sails 14 km on a bearing of 080°, then 9 km on a bearing of 190°. How far is it from its starting dock?

From two points A and B on shore, 400 m apart, a kayaker is seen at angles of 71° (at A) and 58° (at B) from the shoreline AB. How far is the kayaker from the shore?

A triangular garden is fenced with sides 12 m, 14 m and 16 m. Find its area using the angle between the 12 m and 14 m sides and Area = ½ab sin C. (Enrichment beyond C1: the height to the 12 m side is 14 sin C, so ½ × base × height = ½(12)(14) sin C.)

A ladder 6.5 m long leans against a wall that is NOT vertical: the ground slopes so the angle between the ground and the wall is 80°. The foot of the ladder is 2.0 m from the base of the wall. How far up the wall does the ladder reach? (Assume the triangle is acute.)

Explain why lookout towers far apart give a better fix on a fire than towers close together.

Challenge

A ship at S is 12 km from lighthouse L on a bearing of 050° from L. A second ship T is 15 km from L on a bearing of 115° from L. How far apart are the ships, and what is the bearing from S to T?

A roof truss is an isosceles triangle with a 9.0 m chord and a peak angle of 70°. Find the rafter length and the pitch angle.

Three radio towers form a triangle: AB = 8.0 km, BC = 9.5 km and AC = 7.2 km. A new tower M will be built on BC, 7.0 km from B. How far is it from A? (Hint: find ∠B first.)

Check your understanding

Quiz

Two towers 15 km apart see a fire at 52° and 61° from the line joining them. What is the angle at the fire?

In Example 1, AF is across from which angle?

A boat sails on 040°, then turns to 150°. The interior angle at the turn is:

A surveyor knows two distances from her position and the angle between them. To find the distance between the two far points she uses:

Which is the BEST final answer to "How far is the fire from tower A?"

Why does a larger baseline usually give a more accurate triangulation?

For parents and teachers

Big idea: Expectation C1.5: solve problems that require the sine law or the cosine law in acute triangles, including real-world applications such as surveying, navigation and building construction.

How to help: The hardest part is the sketch. Ask your teen to read the question aloud, then draw and label as they go. For navigation, check for a north line at every point and a back bearing at each turn.

Common difficulties:

Using a bearing as if it were an interior angle of the triangle.
Pairing a side with the wrong angle (the side across from a vertex never touches it).
Stopping before answering the actual question (for example, finding AF when the question asks for the distance to the road).
For the outdoor activity: stay off roads and keep away from water.