Grade by Grade
C. Trigonometric Functions · Trigonometry of Acute Triangles

Multi-Step Problems: Combining Triangles in 2-D and 3-D

C1.5 C1.4 C1.2About 75 minutes
Learning goals
  • I can break a complicated figure into right triangles and acute triangles and plan the order to solve them.
  • I can combine the sine law, the cosine law and the primary ratios in one problem.
  • I can solve 3-D problems where one triangle lies on the ground and another stands up from it.
  • I can find perimeters and areas of shapes made from triangles.

Words to know

multi-step problem
A problem that needs more than one triangle; the answer from one becomes a given for the next.
area of a triangle (SAS)
Area = ½ab sin C, using two sides and the angle between them.
horizontal angle
An angle measured flat along the ground, between two directions.

Putting it all together

Real measuring jobs rarely fit into one triangle. A surveyor mapping a field with four corners, an engineer checking a tower, a search-and-rescue crew spotting two boats from a cliff: each needs a chain of triangles. This lesson pulls together every tool in the unit. The 3-D problems and the area formula ½ab sin C go a little beyond the curriculum expectations, so treat them as enrichment.

Planning routine:

  1. Sketch the whole situation. For 3-D, draw the ground as a slanted shape and heights straight up.
  2. List the triangles. For each one, write what is known.
  3. Find a triangle you can solve now. Solve only what you need from it.
  4. Pass that answer to the next triangle. Keep full calculator values.
Worked example

Example 1: Height of a tower you can't reach

Surveyors at A and B, 60 m apart on level ground, cannot walk to the base F of a tower. At A, the horizontal angle between B and F is ∠FAB = 62°; at B it is ∠FBA = 71°. From A, the angle of elevation to the top T of the tower is 34°. How tall is the tower?

  1. Triangle 1 lies on the ground: △ABF with ∠A = 62°, ∠B = 71°, AB = 60 m (ASA).
  2. ∠F = 180° − 62° − 71° = 47°. AB is across from ∠F.
  3. AF is across from ∠B: AF = 60 sin 71° ÷ sin 47° ≈ 77.57 m.
  4. Triangle 2 stands up: △AFT, right angle at F (the tower is vertical).
  5. h = AF tan 34° ≈ 77.57 × 0.6745 ≈ 52.3 m.

The tower is about 52 m tall.

62°71°34°ABFTAB = 60 mh1. Ground triangle ABF:sine law → AF2. Upright triangle AFT:tangent → h
Example 1. The shaded ground triangle ABF is solved with the sine law; then the upright right triangle AFT gives the height. The right angle at F does not look like 90° in a 3-D sketch.
Worked example

Example 2: A four-sided field

A field ABCD is split by the diagonal AC. In △ABC, AB = 85 m, BC = 70 m and ∠ABC = 64°. In △ACD, ∠CAD = 58° and ∠ACD = 60°. Find the perimeter of the field and its area.

  1. △ABC (SAS): AC² = 85² + 70² − 2(85)(70) cos 64° ≈ 12 125 − 5216.62 ≈ 6908.38, so AC ≈ 83.12 m.
  2. △ACD (ASA): ∠ADC = 180° − 58° − 60° = 62°. AC is across from ∠D.
  3. CD (across from 58°) = 83.12 sin 58° ÷ sin 62° ≈ 79.83 m.
  4. AD (across from 60°) = 83.12 sin 60° ÷ sin 62° ≈ 81.52 m.
  5. Perimeter = 85 + 70 + 79.83 + 81.52 ≈ 316.4 m.
  6. Area of △ABC = ½(85)(70) sin 64° ≈ 2674 m². Area of △ACD = ½(AC)(AD) sin 58° ≈ 2873 m².
  7. Total area ≈ 5547 m².

The perimeter is about 316 m and the area is about 5547 m² (about 0.55 hectares).

64°58°60°ABCD85 m70 mACStep 1: △ABC is SAS→ cosine law for ACStep 2: △ACD is ASA→ sine law for CD, AD
Example 2, to scale. The diagonal AC is the shared side: found with the cosine law in the first triangle, used with the sine law in the second.

Enrichment: area from two sides and the angle between them

This formula goes beyond the C1 expectations, but it uses only right-triangle ideas from this unit. If you know two sides a and b of a triangle and the angle C between them, the height to side a is b sin C (a right-triangle fact). So:

Area = ½ab sin C

This is handy in multi-step problems, because after solving a triangle you usually know an SAS combination.

Worked example

Example 3: Two boats seen from a cliff

From the top of a 60 m cliff, the angle of depression to boat P is 18° and to boat Q is 24°. At the base of the cliff, the horizontal angle between the directions to the two boats is 50°. How far apart are the boats?

  1. Two upright right triangles give the distances from the base of the cliff (angles of depression equal angles of elevation).
  2. Base to P: 60 ÷ tan 18° ≈ 184.66 m. Base to Q: 60 ÷ tan 24° ≈ 134.76 m.
  3. The ground triangle has these two sides and the 50° angle between them: SAS, cosine law.
  4. PQ² = 184.66² + 134.76² − 2(184.66)(134.76) cos 50° ≈ 20269
  5. PQ ≈ 142.4 m

The boats are about 142 m apart.

50°top of cliffbase60 mboat Pboat QPQ = ?18°24°
Example 3. The angles of elevation from the boats (18° and 24°) equal the angles of depression from the cliff top. Two upright right triangles give the ground distances to each boat. The shaded ground triangle is then solved with the cosine law.
Just for fun

A funny-but-true fact: a triangle is the only polygon that cannot change shape without changing a side length. That is why bridges, cranes and roof trusses are full of triangles, and why a wobbly bookshelf is fixed with a diagonal brace.

Hands-on activity

Map a room or a yard with triangles

You need: a measuring tape, paper, a protractor and a calculator. Pick a four-sided area: a room, a deck, a garden bed or a section of yard.

  1. Label the corners A, B, C and D. Measure AB, BC and the angle at B (use the protractor on a large piece of cardboard placed in the corner, or a phone angle app).
  2. Measure the angles ∠CAD and ∠ACD from A and C, sighting along the walls or edges.
  3. Calculate AC, CD, AD, the perimeter and the area, as in Example 2.
  4. Measure CD and AD with the tape to check.

What to look for: checked lengths within about 3%. Most rooms have 90° corners, so you can also check your answers with the Pythagorean theorem for a rectangular room.

Your turn

1

From points A and B, 80 m apart, the horizontal angles to the base F of a radio mast are ∠FAB = 57° and ∠FBA = 68°. From B, the angle of elevation to the top is 30°. Find the height of the mast.

2

In quadrilateral PQRS, PQ = 12 cm, QR = 15 cm and ∠PQR = 70°. Find the diagonal PR.

3

Continue: in △PRS, ∠RPS = 54° and ∠PRS = 66°. Find PS and RS.

4

Find the area of a triangle with sides 9 m and 11 m and a 52° angle between them.

5

From the top of a 45 m lighthouse, the angles of depression to two buoys are 15° and 21°. The horizontal angle between the buoys at the base of the lighthouse is 60°. How far apart are the buoys?

6

Two right triangles and an acute triangle: a 20 m flagpole is seen from A at 40° elevation and from B at 32° elevation. A and B are on level ground and the horizontal angle ∠AFB at the base of the pole is 75°. Find AB.

More practice: warm-up

1

Write the area formula for a triangle using two sides and the angle between them.

2

A vertical tower makes what angle with the level ground?

3

Find ½(10)(8) sin 30°.

4

A field is split into two triangles with areas 1250 m² and 1730 m². Find the total area.

5

Two angles of a ground triangle are 50° and 75°. Find the third.

More practice: on your own

1

A triangular plot has sides 40 m and 52 m with a 66° angle between them. Find its third side, its perimeter and its area.

2

A kite (the four-sided shape) has two sides of 30 cm meeting at a 70° angle and two sides of 45 cm. Find the diagonal joining the ends of the 30 cm sides, then use it to find the angle between the two 45 cm sides.

3

In Example 1, Marco used AB = 60 m directly with tan 34° to find the height. Explain the mistake.

4

A hiker at H sees two peaks. Peak X is 4.2 km away and peak Y is 5.5 km away (horizontal distances), and the horizontal angle between them at H is 48°. How far apart are the peaks horizontally?

5

Two observers A and B are 300 m apart. A balloon is directly above point P on the ground. ∠PAB = 54° and ∠PBA = 63°. The angle of elevation from A to the balloon is 22°. How high is the balloon?

Challenge

1

In a regular pentagon with 10 cm sides, the two diagonals from one vertex and the side opposite that vertex form an acute isosceles triangle with angles 36°, 72° and 72°. The 10 cm side is across from the 36° angle. Find the length of a diagonal.

2

From the top of a 70 m building, the angles of depression to two cars on level ground are 25° and 35°. The horizontal angle between the cars at the base is 55°. The cars then drive straight toward each other at the same speed. How far does each drive before they meet?

3

A triangular prism-shaped tent has a ridge pole 2.5 m long. Each end is an acute isosceles triangle with two 2.0 m sides meeting at the top at 64°. Find the width of the tent floor and the area of canvas needed for the two sloping sides.

Check your understanding

Quiz

1.

In Example 1, why is the sine law used before tangent?

2.

Area of a triangle with sides 9 and 11 and a 52° angle between them:

3.

A diagonal splits a quadrilateral into two triangles. The diagonal is:

4.

From the top of a 60 m cliff, the angle of depression to a boat is 18°. Its distance from the base is:

5.

Two ground distances from a point are 185 m and 135 m with a 50° angle between them. The distance between the far ends uses:

6.

Which habit avoids small errors in multi-step problems?

For parents and teachers

Big idea: Expectation C1.5 (problems with the sine law or the cosine law, including real-world applications) combined with C1.2 (two-triangle problems) and C1.4 (choosing the law). The 3-D examples go a little beyond the curriculum's two-dimensional wording and can be treated as enrichment, as can the area formula ½ab sin C.

How to help: Ask your teen to write a plan in words before calculating, such as "Step 1: ground triangle, sine law for AF. Step 2: upright triangle, tangent for h." A shoebox or a pencil standing on a book helps make 3-D sketches real.

Common difficulties:

  • Using a distance from one triangle in the wrong triangle (Marco's mistake).
  • Losing track of which angle is horizontal and which is vertical in 3-D.
  • Rounding the shared side too early.