D. Trigonometric Functions · Trigonometric Ratios, Identities and Laws

Problems in Three Dimensions: Oblique Triangles

D1.7About 75 minutes

Learning goals

I can tell horizontal angles (measured on the ground) from angles of elevation (measured upward).
I can solve 3-D problems that combine right triangles with the sine law or the cosine law.
I can find angles between lines in a box or a pyramid using the cosine law.
I can pose and solve my own 3-D problem.

Words to know

horizontal angle
An angle measured in a level plane, such as the angle on the ground between the directions to two landmarks.

vertical plane
A plane that stands straight up, like a wall. Angles of elevation are measured in vertical planes.

face diagonal
A diagonal of one rectangular face of a box.

apex
The top vertex of a pyramid or cone.

When the 3-D triangle isn't a right triangle

In lesson 12 every triangle you used was a right triangle. But the ground triangle between two observers and the foot of a tower doesn't have to be right-angled, and neither does the triangle formed by two diagonals of a box. In this lesson you will combine right triangles and oblique triangles in the same problem.

Two kinds of angles

Horizontal angles are measured in the ground plane, for example ∠CAB between the directions from A to B and from A to the tower's foot C. They belong to the ground triangle.
Angles of elevation (or depression) are measured upward from the horizontal, in a vertical plane. They belong to a vertical right triangle.

Never mix them in the same triangle. Label each angle on your sketch with the triangle it belongs to.

Worked example

Example 1: A tower you can't reach

Surveyors at A and B, 200 m apart on level ground, want the height of a tower whose foot C is across a river. They measure the horizontal angles ∠CAB = 50° and ∠CBA = 70°. From A, the angle of elevation of the top T is 30°. Find the tower's height.

Ground △ABC (oblique, AAS): ∠ACB = 180° − 50° − 70° = 60°.
Sine law: AC/sin 70° = 200/sin 60°, so AC = 200 sin 70° ÷ sin 60° ≈ 217.01 m.
Vertical △ACT is right-angled at C: h = AC tan 30° ≈ 217.01 × 0.5774 ≈ 125.3 m.

The tower is about 125.3 m tall.

The ground triangle ABC (oblique) gives AC. The vertical right triangle ACT then gives the height. (Sketch, not to scale.)

Worked example

Example 2: The angle between two face diagonals

A box is 6 cm long, 4 cm deep and 3 cm high. From corner O, a diagonal OP runs across the front face and a diagonal OQ runs across the side face. Find the angle θ between them.

Each diagonal is the hypotenuse of a right triangle on its face: OP = √(6² + 3²) = √45 ≈ 6.708 cm and OQ = √(4² + 3²) = 5 cm.
P and Q are both on the top face, so PQ is a diagonal of the top face: PQ = √(6² + 4²) = √52 ≈ 7.211 cm.
△OPQ is oblique. SSS, so use the cosine law: cos θ = (45 + 25 − 52) ÷ (2 × √45 × 5) = 18 ÷ (10√45) ≈ 0.2683.
θ = cos⁻¹(0.2683) ≈ 74.4°.

θ ≈ 74.4°. Three right triangles gave the three sides of an oblique triangle.

Each side of the shaded triangle OPQ is a diagonal of a different face of the box. Using squares of the lengths (45, 25, 52) avoids rounding.

Worked example

Example 3: A hot-air balloon

A hot-air balloon H floats 800 m directly above point G on flat farmland. From A, the angle of elevation of the balloon is 25°; from B, it is 38°. On a map, the angle ∠AGB is 110°. How far apart are A and B?

Vertical right △AGH: AG = 800 ÷ tan 25° ≈ 1715.6 m.
Vertical right △BGH: BG = 800 ÷ tan 38° ≈ 1024.0 m.
Ground △AGB is oblique, with the 110° angle between AG and BG (SAS): AB² = 1715.6² + 1024.0² − 2(1715.6)(1024.0) cos 110° ≈ 5 193 400.
AB ≈ 2279 m.

A and B are about 2.28 km apart.

Two vertical right triangles give AG and BG. The ground triangle AGB, with its 110° angle, is solved with the cosine law. (Sketch, not to scale.)

Worked example

Example 4: The angle at the top of a tent

The pyramid tent from lesson 12 has a 3 m square base, and each sloping edge is √8.5 ≈ 2.92 m. Find the angle at the apex between two edges that end at the same base edge.

The triangle formed by two sloping edges and one base edge has sides √8.5, √8.5 and 3.
cos θ = (8.5 + 8.5 − 9) ÷ (2 × 8.5) = 8/17 ≈ 0.4706.
θ ≈ 61.9°.

About 61.9°. (Using the exact value √8.5 means the squares are exactly 8.5.)

Just for fun

Surveyors really do measure things they can't touch, like the height of a tower across a river, by standing in two places and measuring angles. It is one of the few jobs where "I'll just stand over here and look at it" is the correct method.

Hands-on activity

Angles inside a box

You need: a closed box (a shoebox or tissue box), a ruler, three pieces of string or thin strips of paper, tape, a protractor and a calculator.

Measure the box: length l, depth w and height h.
Pick a bottom front corner O. Tape a string from O across the front face to the opposite corner P, and another from O across the side face to the opposite corner Q. Tape a third string across the top from P to Q.
Measure the three strings. Compare with √(l² + h²), √(w² + h²) and √(l² + w²).
Use the cosine law to calculate the angle at O between the first two strings.
Hold the protractor as close to the corner as you can and estimate the angle between the strings.

What to look for: string lengths within a few millimetres of the predictions, and the measured angle within about 5° of the calculated one. (For a 6 × 4 × 3 box shape, the angle is about 74°.)

Your turn

A and B are 150 m apart on level ground. The foot C of a tower makes horizontal angles ∠CAB = 62° and ∠CBA = 48°. From B, the angle of elevation of the top is 35°. Find the height of the tower.

A box is 8 cm by 6 cm by 4 cm. Find the angle at a corner between the diagonals of the 8 × 4 face and the 6 × 4 face.

A square-based pyramid has base edges of 10 m and sloping edges of 9 m. Find the angle at the apex between two edges that meet the same base edge.

A balloon is 500 m above point G. From A the angle of elevation is 30°, from B it is 45°, and ∠AGB = 80°. Find AB.

In a cube, two face diagonals start at the same corner on different faces. What is the angle between them? Explain without a calculator.

More practice: warm-up

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

Find the diagonal of a 6 by 8 face.

Write the cosine law solved for cos A.

Is an angle of elevation a horizontal angle or a vertical one?

A triangle has three equal sides. What is each angle?

More practice: on your own

Two hikers A and B are 1.2 km apart on level ground. The point F directly below a hilltop lookout makes horizontal angles ∠FAB = 40° and ∠FBA = 75°. From A, the angle of elevation of the lookout is 4°. How high is the lookout above the hikers?

A box is 5 cm by 3 cm by 2 cm. Find the angle at a corner between the diagonals of the 5 × 2 face and the 3 × 2 face.

A pyramid tent has a 2.4 m square base and 2.0 m sloping edges. Find the angle at the apex between two edges that meet the same base edge.

A drone hovers 120 m above G. From A the angle of elevation is 20° and from B it is 32°. A and B are 400 m apart. Find the horizontal angle ∠AGB.

Jordan solved Example 1 by putting the 30° angle of elevation into the ground triangle ABC. Explain the error.

Challenge

From A, the angle of elevation of the top of a tower is 20°; from B it is 30°. A and B are 500 m apart, and the horizontal angle ∠AFB at the tower's foot is 120°. Find the height h.

Find the angle between two space diagonals of a cube.

Pose a 3-D problem about your own room that needs both a right triangle and the cosine law. Solve it.

Check your understanding

Quiz

In Example 1, which triangle contains the 30° angle of elevation?

Surveyors know AB = 200 m, ∠CAB = 50° and ∠CBA = 70° on the ground. How do they find AC?

In a 6 × 4 × 3 box, what is the diagonal of the 6 × 4 face?

A triangle has sides √45, 5 and √52. What is cos θ for the angle between √45 and 5?

A balloon is 800 m above G. From A, the angle of elevation is 25°. What is AG?

Two face diagonals of a cube meet at a corner. The angle between them is:

For parents and teachers

Big idea (D1.7): many 3-D problems mix right triangles (usually vertical ones with angles of elevation) and oblique triangles (often on the ground, or inside a box). Students solve them one triangle at a time, with the sine law or cosine law where there is no right angle.

How to help: Ask your teen to colour the ground triangle one colour and each vertical triangle another, and to label each angle with the triangle it belongs to. Use squared lengths (like 45 instead of 6.708) where possible to avoid rounding errors.

Common difficulties:

Putting an angle of elevation into the ground triangle (or a ground angle into a vertical triangle).
Assuming the ground triangle has a right angle when it doesn't.
Rounding the first triangle's answer too early.