C1.2 Solve two-dimensional problems involving two right triangles

Ontario Grade 11 Mathematics

Trigonometric Functions

Use right-triangle trigonometry to connect two parts of one situation

Some two-dimensional situations contain more than one right triangle. Each triangle can describe part of the situation, but the triangles are often connected by a shared side or by lengths that differ by a known amount. The key is to label both triangles, use the information given, and connect the equations. This lesson reviews the trigonometry needed for one right triangle, then applies it to a situation with two.

What you will learn

Identify two right triangles in a two-dimensional situation.
Choose a trigonometric ratio that connects known and unknown sides.
Use a shared length or a relationship between lengths to form and solve an equation.
Check that an answer fits the situation and uses appropriate units.

1. Prerequisite bridge: one right triangle

A right triangle has one angle measuring

90^\circ

. The longest side, opposite the right angle, is the hypotenuse. For a chosen acute angle, the opposite side is across from that angle, and the adjacent side touches it but is not the hypotenuse.

The three basic trigonometric ratios connect an acute angle to two side lengths. Use sine for opposite over hypotenuse, cosine for adjacent over hypotenuse, and tangent for opposite over adjacent. A calculator must be in degree mode when the angles are given in degrees.

For example, if a right triangle has a

30^\circ

angle and an adjacent side of

8

units, tangent is useful for finding the opposite side: it relates those two sides directly. You do not need to find the hypotenuse first.

\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}, \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}, \tan\theta=\frac{\text{opposite}}{\text{adjacent}}

Mark the angle you are using before deciding which sides are opposite and adjacent.
The hypotenuse is always opposite the right angle.
Use the ratio that includes the side you know and the side you need.

2. Plain language: connect the two triangles

A two-dimensional problem involving two right triangles asks you to use both triangles to determine a quantity. The triangles may share a side, such as the height of a building. They may also use sides whose lengths are related, such as two observation points separated by a known distance.

A useful plan is to draw the situation, mark each right angle, and label the known measurements. Then identify a trigonometric ratio for each triangle. If both triangles contain the same unknown side, set the two expressions for that side equal. If their sides differ by a known distance, write that relationship into an equation.

An angle of elevation is measured upward from a horizontal line to an object. An angle of depression is measured downward from a horizontal line. These angles help describe a triangle, but the horizontal and vertical sides still meet at a right angle.

One triangle may provide an equation that the other triangle also uses.
A known distance between points can connect the triangles even when they do not share a side.
Keep units consistent, and state the unit with the final measurement.

3. Represent the situation before calculating

A sketch does not need to be drawn to scale. It needs to show which lengths and angles belong together. For each triangle, make a small side-and-angle label list. This prevents a common error: using the opposite side from one triangle with the angle from the other.

Suppose two people stand on level ground in line with a tower. Each person looks up at the top. The ground, tower, and line of sight make a right triangle for each person. Both triangles have the tower height as their vertical side. The farther person has a longer horizontal distance to the tower, so that distance includes the gap between the two people.

After sketching, translate the labels into equations. For an angle of elevation, the vertical height is opposite the angle and the horizontal distance is adjacent. Tangent therefore connects the angle, height, and horizontal distance. The equations can be rearranged or combined to find the unknown measurements.

\tan\theta=\frac{\text{vertical height}}{\text{horizontal distance}}

A sketch organizes information; it is not a measurement tool.
Name each horizontal distance clearly, especially when one includes a separation between points.
Use a shared quantity, such as height, to connect the equations.

4. Apply the method and check the result

In the worked example, the two observation points are on the same straight, level line from the tower. The near point is

x

metres from the tower. The far point is

x+18

metres from it. The tower height is the same in both triangles, so each angle creates a different expression for that height.

The equation from the nearer point gives a greater height expression for the same horizontal distance variable. The equation from the farther point uses a longer distance and a smaller angle. Equating these expressions lets us solve for the near distance, then substitute that distance to find the height.

In other settings, the shared quantity may be a side other than height. Always check what the diagram actually shows. Once you have a numerical answer, compare it with the sketch: a farther point should have a smaller angle of elevation to the same top, and a height should be positive.

Write one relationship for each triangle before combining them.
Solve for one unknown at a time when possible.
Round only after the calculation is complete.

Labels for the two observation triangles

Triangle	Angle of elevation	Horizontal distance	Vertical height
Near	$38^\circ$	$x$ m	$h$ m
Far	$24^\circ$	$(x+18)$ m	$h$ m

Worked example

Find a tower’s height from two viewing points

Two people stand on level ground in a straight line from a tower. The nearer person measures an angle of elevation of

38^\circ

to the top. The farther person measures

24^\circ

. The people are

18

m apart. Find the tower’s height to the nearest tenth of a metre.

Define the unknowns
Let $x$ be the horizontal distance from the tower to the nearer person, in metres. Let $h$ be the tower’s height, in metres. The farther person is $18$ m beyond the nearer person, so that person is $x+18$ m from the tower.
Write the nearer-triangle equation
For the nearer person, the tower height is opposite the $38^\circ$ angle and the horizontal distance is adjacent. Tangent relates those sides.
$\tan 38°=\frac{h}{x}$
Write the farther-triangle equation
For the farther person, the same tower height is opposite the $24^\circ$ angle. The adjacent side is now $x+18$ .
$\tan 24°=\frac{h}{x+18}$
Set the height expressions equal
Both equations describe the same height, so rearrange each for $h$ and set the expressions equal. Solve for the nearer distance $x$ .
$x\tan 38°=(x+18)\tan 24°$
Calculate the distance and height
Using degree mode gives $x\approx23.8$ m. Substituting this value into the nearer-triangle equation gives the height. Keep extra calculator digits until the final rounding.
$h\approx 23.8\tan 38°\approx18.6$

Answer: The tower is approximately

18.6

m tall.

Check: The farther point is about

41.8

m from the tower, more than the nearer point’s

23.8

m. A smaller angle at the farther point is reasonable. Substitution gives approximately

18.6

m from either triangle.

Common mistakes and how to avoid them

Using the

18

m separation as the far person’s distance from the tower.

Correction: The

18

m is the distance between the people. The far person’s distance from the tower is

x+18

m.

Using sine or cosine without checking which sides are involved.

Correction: Identify opposite, adjacent, and hypotenuse relative to the chosen angle. Here, the known angle relates the vertical height to the horizontal distance, so use tangent.

Treating the two angles as if they describe one triangle.

Correction: Each angle belongs to a different observation point and a different triangle. Write one equation for each.

Rounding intermediate values too soon or reporting no units.

Correction: Keep calculator precision until the last step, then round the height and include metres.

Lesson summary

Draw and label both right triangles in the situation.
Choose a trigonometric ratio using the sides and angle in each triangle.
Connect the equations with a shared side or a known distance relationship.
Solve, round at the end, include units, and check that the result makes sense.

Check your understanding

Question 1

Two people view the top of the same vertical pole from level ground. The nearer person is

x

metres from the pole, and the farther person is

x+12

metres away. Which equation matches the two equal expressions for the pole’s height if their angles of elevation are

40^\circ

and

25^\circ

, respectively?

$x\tan 40^\circ=(x+12)\tan 25^\circ$
$x\tan 25^\circ=(x+12)\tan 40^\circ$
$\frac{x}{\tan 40^\circ}=\frac{x+12}{\tan 25^\circ}$
$x\tan 40^\circ=(x+12)\tan 40^\circ$

Show answer and explanation

x\tan 40^\circ=(x+12)\tan 25^\circ

For each point, height equals horizontal distance times the tangent of that point’s angle. The pole height is shared, so the nearer expression uses

x

and

40^\circ

, while the farther expression uses

x+12

and

25^\circ

.

Question 2

A learner finds a positive tower height from two triangles, but the farther point has a larger angle of elevation than the nearer point. What should the learner check first?

Whether the farther point’s horizontal distance was written as the near distance plus the separation
Whether the tower height was treated as shared by both triangles
Whether the farther point’s angle and horizontal distance were paired in the same triangle
Whether the final height was rounded to the nearest tenth

Show answer and explanation

Whether the farther point’s angle and horizontal distance were paired in the same triangle

Each angle must be paired with the horizontal distance from its own observation point. A mismatch can reverse the expected relationship between distance and angle. For points on the same level line viewing the same top, the farther point should have the smaller angle.

Key terms

Right triangle: A triangle with one angle measuring $90^\circ$ .
Hypotenuse: The side opposite the right angle. It is the longest side of a right triangle.
Angle of elevation: An angle measured upward from a horizontal line to an object.
Trigonometric ratio: A ratio that relates an acute angle in a right triangle to two side lengths.

Continue through MCF3M

View the complete Ontario Grade 11 Mathematics learning path

About this lesson

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Mathematics (MCF3M), expectation C1.2. It is a study resource, not an official curriculum publication.

Official curriculum reference

Report a correction or ask a question

C1.2 · Solve two-dimensional problems involving two right triangles