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

Two Right Triangles: Heights and Distances

C1.2 C1.1About 75 minutes
Learning goals
  • I can recognize when a problem needs two right triangles and sketch them.
  • I can decide whether two distances should be added or subtracted.
  • I can use a shared side to connect the two triangles, either step by step or with an equation.
  • I can solve height and distance problems from two observation points.

Words to know

shared (common) side
A side that belongs to both right triangles. It links them: find it in one triangle, then use it in the other.
observation point
A place where an angle is measured, such as a surveyor's instrument or a window.
system of equations
Two or more equations with the same unknowns, solved together.

When one triangle is not enough

A lot of real measuring needs two right triangles. You might stand on either side of a tower, step closer and measure again, or look up and down from a window. The trick is to find what the two triangles share. Expectation C1.2 is all about these two-triangle problems in two dimensions.

In this lesson the two triangles share a vertical side (a height). In the next lesson they share other sides, and you will add bearings and surveying.

Object betweenobservers: ADDBoth on one side:SUBTRACTFrom a window:split at horizontal
Three common set-ups. Left: the object is between the observers, so the two ground distances add. Middle: both observers are on the same side, so one distance is part of the other. Right: from a window, the horizontal line of sight splits the far building into two parts.

Plan before you calculate

  1. Draw both triangles in one sketch. Mark every right angle.
  2. Mark the shared side.
  3. Ask: is the length I want a sum or a difference of two pieces?
  4. Find the shared side in the triangle where you have enough information, then switch triangles.
  5. Keep unrounded values in your calculator between steps.
Worked example

Example 1: Observers on opposite sides

Two friends stand on opposite sides of a 40 m tall radio tower, in line with its base. The angles of elevation to the top are 35° and 50°. How far apart are the friends?

  1. Each friend makes a right triangle with the tower. The tower (40 m) is the shared side, opposite each angle.
  2. Friend 1: distance = 40 ÷ tan 35° ≈ 57.13 m.
  3. Friend 2: distance = 40 ÷ tan 50° ≈ 33.56 m.
  4. They are on opposite sides, so add: 57.13 + 33.56 ≈ 90.7 m.

The friends are about 90.7 m apart.

Worked example

Example 2: Both observations from one side

A surveyor at A measures the angle of elevation to the top T of a cliff as 32°. She walks 50 m straight toward the cliff to B and measures 47°. Find the height h of the cliff. (Ignore the height of the instrument.)

  1. Let D be the base of the cliff and x = BD. Then AD = x + 50.
  2. Neither triangle has a known side yet, so write h two ways:
  3. In △BDT: tan 47° = h/x, so h = x tan 47°.
  4. In △ADT: tan 32° = h/(x + 50), so h = (x + 50) tan 32°.
  5. Set them equal: x tan 47° = x tan 32° + 50 tan 32°.
  6. Collect x terms: x (tan 47° − tan 32°) = 50 tan 32°.
  7. x = 50 tan 32° ÷ (tan 47° − tan 32°) ≈ 69.82 m.
  8. h = x tan 47° ≈ 74.9 m. Check with the other triangle: (69.82 + 50) tan 32° ≈ 74.9 m. ✓

The cliff is about 75 m high.

32°47°ABDT50 mxh△BDT: h = x tan 47°△ADT: h = (x + 50) tan 32°
Example 2, drawn to scale. The cliff height h is the shared side of the two right triangles. Because the surveyor moved 50 m closer, the angle of elevation grew from 32° to 47°.
Worked example

Example 3: Up and down from a window

Jenna looks out a window 15 m above the ground. Across the street, 28 m away, is an office building. The angle of elevation to the top of the office building is 38°. Find the angle of depression to its base and the height of the office building.

  1. Draw a horizontal line from the window to the office building. It splits the building into an upper piece and a lower piece of 15 m.
  2. Lower triangle: tan(depression) = 15/28, so the angle of depression is tan⁻¹(15/28) ≈ 28°.
  3. Upper triangle: upper piece = 28 tan 38° ≈ 21.88 m.
  4. Height = upper + lower = 21.88 + 15 ≈ 36.9 m.

The angle of depression is about 28° and the office building is about 36.9 m tall.

Just for fun

Riddle: what do you call two right triangles that share a side?

Good neighbours. When one of them is stuck, the other always lends it a hand... well, a height.

Hands-on activity

Two-angle height of your house

You need: the clinometer from the last lesson (protractor, straw, string and weight), a tape measure and a partner.

  1. Pick a tall object whose base you can't easily reach, or pretend you can't (for example, the peak of your roof seen across a yard).
  2. Measure the angle of elevation to the top from a first point. Walk a measured distance (say 5 m) straight toward the object and measure again.
  3. Use the method of Example 2 to find the height above eye level, then add your eye height.
  4. If you can reach the base, measure the distance to check your value of x.

What to look for: the second angle must be larger than the first. The farther apart the two points, the more reliable the answer. Small angle errors make big height errors when the two angles are close together.

Your turn

1

Two people stand on opposite sides of a 25 m flagpole, in line with it. Their angles of elevation to the top are 40° and 58°. How far apart are they?

2

From a point 60 m from the base of a building, the angles of elevation to the bottom and top of a flagpole on the roof are 38° and 43°. How tall is the flagpole?

3

From a 30 m lookout tower, the angles of depression to two campsites in a straight line on the same side of the tower are 20° and 35°. How far apart are the campsites?

4

From a window 9 m up, the angle of elevation to the top of a building across the street is 41° and the angle of depression to its base is 23°. How tall is the building?

5

From point A the angle of elevation to the top of a tower is 28°. From point B, 40 m closer (in line), it is 41°. Find the height of the tower.

6

In Example 1, which friend is closer to the tower, and how can you tell from the angles alone?

More practice: warm-up

1

Two ground distances are 34 m and 21 m on opposite sides of a pole. What is the total distance?

2

Find x if 30 = x tan 40°.

3

Solve 2x = 0.8x + 12.

4

A height is 18 m and an angle of elevation is 45°. What is the horizontal distance?

5

From a window, the horizontal splits a building into pieces of 12 m and 9.5 m. How tall is the building?

More practice: on your own

1

A hot-air balloon hovers between two towns 12 km apart, directly above the line joining them. The angles of depression from the balloon to the towns are 9° and 14°. How high is the balloon? (Hint: write both horizontal distances in terms of h.)

2

Two buildings are 42 m apart. From the roof of the shorter one (20 m tall), the angle of elevation to the roof of the taller one is 26°. How tall is the taller building?

3

Kai solved Example 2 by finding h = 50 tan 47° − 50 tan 32°. Explain why this is wrong.

4

A 22 m tree is on the far bank of a river. From the near bank, directly opposite, the angle of elevation to its top is 31°. From a point 25 m farther back (in line), what is the angle of elevation?

5

A satellite dish is mounted on a pole on a roof. From 18 m away from the house, the angle of elevation is 34° to the roof edge where the pole stands and 39° to the top of the pole. The pole stands directly above the roof edge. How tall is the pole?

Challenge

1

Show that in Example 2 the height is h = d tan α tan β ÷ (tan β − tan α), where d is the distance walked, α is the first (smaller) angle and β is the second.

2

Use the formula from the last question to check Example 2.

3

From the top of a 50 m building, the angles of depression to the top and bottom of a shorter building are 25° and 48°. Find the height of the shorter building.

Check your understanding

Quiz

1.

Two observers are on opposite sides of a tower, in line with it. The total distance between them is:

2.

Two angles of elevation to the same tower top are measured from the same side. Which point is closer to the tower?

3.

Two friends on opposite sides of a 40 m tower see the top at 35° and 50°. About how far apart are they?

4.

In Example 2 (32° and 47°, 50 m apart), why did we write h two ways?

5.

From a window 15 m up, a building 28 m away is seen. The angle of depression to its base is about:

6.

Which is the BEST first step in any two-triangle problem?

For parents and teachers

Big idea: Expectation C1.2 asks students to solve problems involving two right triangles in two dimensions. This lesson covers the most common type: two triangles sharing a vertical side (a height).

How to help: Ask, "What do the two triangles share?" and "Should these distances add or subtract?" before any calculation. If your teen is stuck on Example 2, point out that neither triangle can be solved alone, so they need an equation.

Common difficulties:

  • Using the distance between the two observers as if it were the distance to the object.
  • Adding when they should subtract (or the reverse).
  • Rounding the shared side in the middle and getting a slightly different final answer.
  • Algebra: collecting the x terms when solving x tan 47° = (x + 50) tan 32°.