C. Trigonometric Functions · Trigonometry of Acute Triangles

Two Right Triangles: Shared Sides, Navigation and Surveying

C1.2 C1.1About 75 minutes

Learning goals

I can solve problems where two right triangles share a side that is not vertical, such as a diagonal or an altitude.
I can split a non-right triangle into two right triangles with an altitude.
I can solve navigation problems with two legs and find a bearing back to the start.
I can find a distance across a river using two angles and an equation.

Words to know

altitude
A segment from a vertex perpendicular to the opposite side. It splits a triangle into two right triangles.

diagonal
A segment joining two corners of a polygon that are not next to each other.

leg (of a trip)
One straight part of a journey, travelled in one direction.

Building bridges between triangles

In the last lesson the two right triangles shared a height. Now the shared side can be anything: an altitude inside a triangle, a diagonal across a garden, or the direct line back to a dock. The strategy is the same:

Find the triangle you can solve now.
Use it to find the shared side.
Move to the second triangle.

An altitude makes two right triangles

Any triangle can be cut into two right triangles by an altitude: a line from one vertex that meets the opposite side at 90°. This trick lets you work with triangles that have no right angle using only SOH CAH TOA. (It is also exactly how the sine law is built, as you will see in a later lesson.)

Example 1, to scale. The altitude AD = 8 cm is the shared side of right triangles ABD and ACD. Each piece of the base comes from a tangent.

Worked example

Example 1: Using an altitude

In △ABC, ∠B = 40°, ∠C = 55° and the altitude from A to BC is AD = 8.0 cm. Find BC, AB and AC.

In right △ABD: tan 40° = 8.0/BD, so BD = 8.0 ÷ tan 40° ≈ 9.53 cm.
In right △ACD: DC = 8.0 ÷ tan 55° ≈ 5.60 cm.
BC = BD + DC ≈ 15.1 cm.
AB is the hypotenuse of △ABD: AB = 8.0 ÷ sin 40° ≈ 12.4 cm.
AC is the hypotenuse of △ACD: AC = 8.0 ÷ sin 55° ≈ 9.8 cm.

BC ≈ 15.1 cm, AB ≈ 12.4 cm, AC ≈ 9.8 cm

Worked example

Example 2: A shared diagonal

Two triangular garden beds share a side AC. In the first bed, ∠ABC = 90°, AB = 12 m and ∠BAC = 38°. In the second bed, ∠ACD = 90° and CD = 7.0 m. Find the length AD and the angle ∠CAD.

First triangle: AB is adjacent to 38° and AC is the hypotenuse: AC = 12 ÷ cos 38° ≈ 15.228 m. (Keep this value in your calculator.)
Second triangle: AC and CD are the legs, AD is the hypotenuse.
AD = √(15.228² + 7.0²) ≈ 16.8 m.
∠CAD = tan⁻¹(7.0 ÷ 15.228) ≈ 25°.

AD ≈ 16.8 m and ∠CAD ≈ 25°

Worked example

Example 3: Navigation with a right-angle turn

A boat leaves a dock and travels 20 km on a bearing of 030°. It then turns and travels 15 km on a bearing of 120° to an island. How far is the island from the dock, and what is its bearing from the dock?

Draw north at the dock and at the turning point. The change in heading is 120° − 30° = 90°, so the two legs meet at a right angle.
Direct distance = √(20² + 15²) = √625 = 25 km.
At the dock, the angle between the first leg and the direct line is tan⁻¹(15/20) ≈ 36.9°.
The first leg is at 030°, and the island is further clockwise, so its bearing is 030° + 36.9° ≈ 066.9°.

The island is 25 km from the dock on a bearing of about 067°.

Example 3. A north arrow is drawn at each point. Because the heading changes by exactly 90°, the two legs and the direct route form a right triangle.

Worked example

Example 4: How wide is the river?

Surveyors mark points A and B, 80 m apart, along a straight riverbank. A tree T stands on the far bank. ∠TAB = 50° and ∠TBA = 62°. Find the width of the river (the perpendicular distance from T to the near bank).

Drop a perpendicular from T to AB, meeting it at D. Let w = TD, the width. Now there are two right triangles sharing w.
△ADT: AD = w ÷ tan 50°. △BDT: DB = w ÷ tan 62°.
AD + DB = 80, so w/tan 50° + w/tan 62° = 80.
Factor: w (1/tan 50° + 1/tan 62°) = 80.
w = 80 ÷ (1/tan 50° + 1/tan 62°) ≈ 80 ÷ 1.3708 ≈ 58.4 m.
Check: AD ≈ 49.0, DB ≈ 31.0, sum = 80.0. ✓

The river is about 58 m wide.

Example 4. The width w is the shared side of two right triangles, and the two pieces of the near bank add to 80 m. (Drawn to scale.)

Just for fun

A navigator's favourite bedtime story? Goldilocks and the Three Bearings: 029° was too far left, 031° was too far right, and 030° was just right.

Hands-on activity

Map a walk with bearings

You need: a phone compass app or a magnetic compass, a measuring tape (or count paces after measuring your pace length), paper and a friend or family member. Do this in a park, a yard or a quiet driveway, not on a road.

Start at a marked spot. Walk a measured distance on one bearing (for example, 15 m on 030°).
Turn so that your new bearing is 90° more (here 120°) and walk another measured distance (for example, 10 m).
Calculate the direct distance back to the start and the bearing from the start to where you are.
Measure the actual distance back with the tape and check the bearing with the compass.

What to look for: distances within a metre or two and bearings within about 5°. Turning exactly 90° is hard by eye; that is often the biggest source of error.

Your turn

In △PQR, the altitude from P to QR is PS = 6.0 cm. ∠Q = 48° and ∠R = 62°. Find QR.

In Example 1, find the area of △ABC.

A hiker walks 6.0 km on a bearing of 070°, then turns and walks 8.0 km on a bearing of 160°. How far is she from the start, and on what bearing?

Points A and B are 100 m apart on one bank of a river. A rock R on the far bank has ∠RAB = 45° and ∠RBA = 60°. Find the width of the river.

A rectangular lot is 30 m by 18 m. A right-triangular flower bed uses the lot's diagonal as one leg and a 5.0 m edge, perpendicular to the diagonal, as the other. Find the hypotenuse of the flower bed.

In right △ABC (∠C = 90°), ∠A = 30° and AB = 10 cm. D is on AC with ∠DBC = 30°. Find AD.

More practice: warm-up

What is an altitude of a triangle?

A boat heads 040°, then turns to 130°. By how many degrees did it turn?

Solve w/2 + w/3 = 10.

Legs are 9 km and 12 km. Find the hypotenuse.

Find 1/tan 45°.

More practice: on your own

A cyclist rides 12 km on a bearing of 200°, then 5.0 km on a bearing of 290°. Find the distance and bearing from the start.

A symmetric roof truss has a span of 9.6 m and a vertical post 2.4 m tall in the middle. Find the angle each rafter makes with the horizontal and the length of each rafter.

An asymmetric roof has a vertical post 3.0 m tall. The left rafter makes 35° with the horizontal and the right rafter makes 50°. Find the total span.

In Example 4, Priya used AD = w tan 50° instead of w ÷ tan 50°. What went wrong?

Two right triangles share hypotenuse PQ = 20 cm. In the first, one acute angle at P is 25°; in the second, one acute angle at P is 40°, and the triangles are on opposite sides of PQ. Find the two legs at P (adjacent to the 25° and 40° angles).

Challenge

In △ABC, ∠B = 40°, ∠C = 55° and BC = 15 cm. Find the altitude AD from A. (Hint: BD = h/tan 40° and DC = h/tan 55°.)

A ship sails 30 km on a bearing of 050°, then 40 km on a bearing of 140°. Find the bearing it must follow to sail straight back to port.

A surveyor wants the distance between two trees, P and Q, on the far side of a pond. From a point A, she finds AP = 120 m due north, and AQ = 160 m due east. Find PQ and ∠APQ.

Check your understanding

Quiz

What does an altitude do to a triangle?

In △ABC, ∠B = 40° and the altitude AD = 8 cm. What is BD?

A boat sails on 030°, then turns to 120°. The angle between the two legs at the turning point is:

A boat goes 20 km on 030°, then 15 km on 120°. The direct distance from the start is:

For the river problem, which equation finds the width w (A and B 80 m apart, angles 50° and 62°)?

In a two-triangle problem you found a shared side of 15.2282... m. What should you do with it?

For parents and teachers

Big idea: More of expectation C1.2 (two right triangles in two dimensions), in navigation and surveying contexts. Splitting a triangle with an altitude is also the foundation for the sine law later in this unit.

How to help: Ask your teen to colour or highlight the shared side in their sketch. For navigation, check that there is a north arrow at the start and at every turn.

Common difficulties:

Forgetting that a 90° change in heading gives a right angle (and that other changes do not).
Adding the angle in the wrong direction when finding a bearing.
Setting up w/tan 50° + w/tan 62° = 80 and then not factoring out w.