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

Solving Real-World Problems with Acute Triangles

3.4 3.3About 75 minutes

Learning goals

I can turn a real-world description into a labelled triangle sketch.
I can choose and apply the sine law or the cosine law to solve real-world problems involving acute triangles.
I can combine the sine or cosine law with right-triangle trigonometry in multi-step problems.
I can communicate a solution clearly with a sketch, equations and a concluding sentence.

Words to know

triangulation
Locating a point by measuring angles to it from two known points.

baseline
The known distance between the two points used for triangulation.

concluding statement
A sentence that answers the question asked, with units.

Triangles hiding everywhere

Forest-fire lookouts, surveyors, sailors, builders and engineers all solve triangles that are not right triangles. In this lesson you will put everything together: read a situation, draw the triangle, choose the right tool, calculate, and explain the answer in words.

Problem-solving routine

Sketch the situation. Label points with capital letters and mark every given length and angle.
Identify the triangle to solve (sometimes there are two).
Choose the tool: right angle → SOH CAH TOA; complete pair → sine law; SAS or SSS → cosine law.
Calculate, keeping unrounded values.
Check: is the triangle acute? Is the longest side opposite the largest angle? Is the answer realistic?
Conclude with a sentence that answers the question, 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 tower B sees it at 61° from that line. How far is the fire from tower A?

Sketch △ABF with AB = 15 km, ∠A = 52° and ∠B = 61°.
Third angle: ∠F = 180° − 52° − 61° = 67°. Now AB (15 km) and ∠F are a complete pair.
AF is opposite ∠B. Sine law: AF/sin 61° = 15/sin 67°.
AF = 15 sin 61° ÷ sin 67° = 14.25... km.
Check: all angles are acute, and AF (opposite 61°) is shorter than AB (opposite 67°). ✓

The fire is about 14.3 km from tower A.

Example 1, drawn to scale. Two angles measured from the ends of a known baseline pin down the fire's position. This is called triangulation.

Worked example

Example 2: Across a lake

A surveyor at C cannot measure straight across a lake from A to B. She measures CA = 350 m, CB = 410 m and ∠ACB = 72°. How long is AB?

SAS (∠C is between CA and CB): cosine law.
AB² = 350² + 410² − 2(350)(410) cos 72° = 122 500 + 168 100 − 287 000 cos 72°.
AB² = 290 600 − 88 688.0... = 201 912.1...
AB ≈ 449.3 m.
Check that the triangle is acute: the longest side is AB, and AB² ≈ 201 912 < 350² + 410² = 290 600. ✓

The lake is about 449 m across from A to B.

Example 2, drawn to scale. The surveyor measures two land distances and the angle between them, then uses the cosine law for the distance across the water.

Worked example

Example 3: A roof truss

A roof truss is a triangle. Its two rafters are 5.2 m and 4.6 m long, and the horizontal span (the bottom chord) is 6.8 m. Find the angle at the peak and the angle each rafter makes with the bottom chord.

SSS: cosine law. The peak angle P is opposite the span (6.8 m).
cos P = (5.2² + 4.6² − 6.8²) ÷ (2 × 5.2 × 4.6) = 1.96 ÷ 47.84 ≈ 0.0410, so ∠P ≈ 87.7°. This is less than 90°, so the truss is an acute triangle.
The angle at the end of the 5.2 m rafter is opposite the 4.6 m rafter: cos = (5.2² + 6.8² − 4.6²) ÷ (2 × 5.2 × 6.8) = 52.12 ÷ 70.72 ≈ 0.7370, so it is about 42.5°.
The last angle: 180° − 87.7° − 42.5° ≈ 49.8° (at the end of the 4.6 m rafter).

Peak ≈ 87.7°; the 5.2 m rafter meets the chord at about 42.5° and the 4.6 m rafter at about 49.8°.

Worked example

Example 4: Two laws in one problem

Two people stand on level ground 1.2 km apart, on opposite sides of a hot-air balloon and in line with it. The angles of elevation to the balloon are 58° and 43°. How high is the balloon?

△ABP: AB = 1.2 km, ∠A = 58°, ∠B = 43°, so ∠P = 180° − 58° − 43° = 79°.
Sine law for AP (opposite ∠B): AP = 1.2 sin 43° ÷ sin 79° = 0.8337... km.
Now drop the height h from P to the ground. In the right triangle at A: h = AP sin 58° = 0.8337... × sin 58° = 0.7070... km.
Convert: 0.707 km ≈ 707 m.

The balloon is about 707 m high.

Example 4, drawn to scale. The sine law finds AP in the acute triangle; then right-triangle trigonometry finds the height h.

Just for fun

Triangulation is so useful that surveyors once mapped huge areas using long chains of triangles linked side to side. Your phone's maps still rely on knowing exactly where things are, which started with people patiently measuring angles.

Hands-on activity

Triangulate a hidden object

You need: a long hallway, yard or park, a measuring tape, two protractors (or one, used twice), string or masking tape, and a partner.

Mark two points A and B on the floor or ground, about 3 to 5 m apart. Measure AB. This is your baseline.
Your partner places an object (a shoe or a ball) somewhere in front of the baseline, so that the triangle looks acute.
Without measuring to the object, stand at A and measure the angle between the line AB and the line to the object (stretch a string along each line and use the protractor). Repeat at B.
Use the sine law to calculate the distance from A to the object.
Now measure that distance with the tape.

What to look for: a labelled sketch, a correct third angle (180° minus the two measured angles), and a calculated distance within about 5% of the measured one. Most error comes from the angle measurements, so take your time with step 3.

Your turn

Two coast guard stations are 20 km apart. From the first, a boat is seen at 48° to the line joining the stations; from the second, at 64°. How far is the boat from the first station?

A hiker walks 6 km on a straight trail, then turns onto another trail and walks 4 km. The angle between the two trails at the turn is 70°. How far is she from her start, in a straight line?

A triangular garden has sides 9 m, 11 m and 12 m. Find its largest angle.

Two radar stations A and B are 50 km apart. A plane P is at ∠PAB = 55° and ∠PBA = 70°. How far is the plane from station A?

A triangular sail has two edges of 4.2 m and 3.8 m that meet at a 64° angle. How long is the third edge?

Points A and B are 300 m apart on level ground, on opposite sides of a tower. The angles of elevation to its top T are 50° from A and 62° from B. How tall is the tower?

More practice: warm-up

You know two angles and the side between them. Which law?

What should be the first step in every word problem in this lesson?

Two angles of a triangle are 52° and 61°. What is the third?

You know two distances and the angle between them. Which law?

Why do we end with a sentence?

More practice: on your own

A surveyor at C measures CA = 500 m, CB = 620 m and ∠ACB = 58° to points on opposite ends of a marsh. Find AB.

A triangular lot has sides 40 m, 46 m and 52 m. Find the largest angle and check that the lot is an acute triangle.

Two forest rangers are 8 km apart. Ranger A sees a campfire at 49° from the line between them, and ranger B sees it at 67°. Which ranger is closer, and how far away is the fire from that ranger?

From a boat, two buoys are 1.8 km and 2.3 km away, and the angle between the lines of sight is 48°. How far apart are the buoys?

Write a complete solution (sketch description, equation, answer, concluding sentence) for: "A triangle has sides 7 m and 9 m with a 48° angle between them. Find the third side."

Challenge

A triangular park has sides 60 m, 70 m and 80 m. Use the cosine law to find one angle, then find the park's area.

An orienteering course goes from control A to control B (400 m), then to control C (500 m). The angle at B between the two legs is 75°. How long is the straight return leg from C to A, and what is the angle at A?

In Example 1, how far is the fire from the line joining the towers (the shortest distance)?

Check your understanding

Quiz

Two towers are 15 km apart. The angles to a fire are 52° at tower A and 61° at tower B. How far is the fire from tower A?

CA = 350 m, CB = 410 m and ∠C = 72°. What is AB?

Two people on opposite sides of a balloon are 1.2 km apart. The angles of elevation are 58° and 43°. About how high is the balloon?

What is the best first step when solving a triangle word problem?

A garden has sides 9 m, 11 m and 12 m. What is its largest angle, to the nearest degree?

You know two angles of a triangle and the side between them. What do you do?

For parents and teachers

Big idea: Expectation 3.4: solving problems involving the measures of sides and angles in acute triangles, often combined with right-triangle trigonometry.

How to help: Read a problem aloud together and have your teen sketch it before any calculation. Ask "What do you know? What do you want? Which tool?" The triangulation activity is a fun way to see the sine law work on real distances.

Common difficulties:

Weak sketches: missing labels, or angles placed at the wrong vertex.
Forgetting to find the third angle before using the sine law.
Finding a side but not answering the question asked (for example, the height in Example 4 needs a second step).
No concluding sentence or missing units.