D. Trigonometric Functions · Trigonometric Ratios, Identities and Laws

Solving Problems in Two Dimensions

D1.6About 75 minutes

Learning goals

I can choose the primary ratios, the sine law or the cosine law to solve a triangle, based on what is given.
I can use true bearings to set up navigation problems.
I can solve multi-step problems that combine a right triangle and an oblique triangle, or two oblique triangles.
I can pose my own problem from a real situation and solve it.

Words to know

true bearing
A direction given as a three-digit angle measured clockwise from north, such as 050° or 160°.

back bearing
The bearing for the return direction: add or subtract 180°. The back bearing of 050° is 230°.

angle of elevation
The angle above the horizontal from an observer's eye up to an object.

pose a problem
Write your own clear question from a situation, with enough information to solve it.

Putting it all together

Now you have the whole toolkit: SOH CAH TOA for right triangles, the sine law (with the ambiguous case) and the cosine law for oblique triangles. Real problems don't come labelled. You have to sketch the situation, find the triangles in it, and pick a tool. This lesson is about making those choices well. The curriculum also asks you to pose problems, so you will write some of your own.

How to choose a tool. Look at what you know about a triangle, then pick the matching law.

Problem-solving routine

Sketch the situation. Mark north lines, horizontal lines and right angles.
Find the triangle that contains the unknown. If it doesn't have enough information, find a neighbouring triangle that does and solve that first.
Choose the tool with the chart above.
Keep full calculator values between steps; round only at the end.
Answer the question asked, with units, in a sentence. Check that the size is reasonable.

True bearings

In Grade 10 you wrote directions like N 40° E (start facing north, turn 40° toward east). Navigation and air traffic usually use true bearings: a three-digit angle measured clockwise from north.

N 40° E is 040°. Due east is 090°. S 20° W is 180° + 20° = 200°.
To find an angle inside a triangle, draw a north line at each turning point. North lines are parallel, so you can use the back bearing (± 180°) or the properties of parallel lines.

Worked example

Example 1: A two-leg voyage

A ferry leaves port P on a bearing of 050° and travels 12 km to point Q. It then turns onto a bearing of 160° and travels 9 km to point R. How far is R from the port, and what is the bearing of R from P?

At Q, the direction back to P has bearing 050° + 180° = 230°. The direction to R has bearing 160°. The angle between them is 230° − 160° = 70°, so ∠PQR = 70°.
SAS, so use the cosine law: PR² = 12² + 9² − 2(12)(9) cos 70° ≈ 225 − 73.88 = 151.12, so PR ≈ 12.3 km.
Angle at P with the sine law: sin(∠QPR) = 9 sin 70° ÷ 12.29 ≈ 0.6880. Side QR = 9 is shorter than PQ = 12, so ∠QPR is acute: ∠QPR ≈ 43.5°.
The ferry turned clockwise (right) at Q, so R is clockwise from Q as seen from P: bearing ≈ 050° + 43.5° = 093.5°.

R is about 12.3 km from port, on a bearing of about 093.5° (just south of due east).

Draw a north line at every point. The back bearing at Q gives the 70° angle inside the triangle.

Worked example

Example 2: Two angles of elevation

From point A, the angle of elevation to the top T of a communications tower is 32°. From point B, 50 m closer to the tower on the same straight line, the angle of elevation is 47°. Find the height of the tower.

△ABT is oblique. ∠ABT = 180° − 47° = 133° (supplementary to the 47° angle). ∠ATB = 180° − 32° − 133° = 15°.
Sine law in △ABT: BT/sin 32° = 50/sin 15°, so BT = 50 sin 32° ÷ sin 15° ≈ 102.37 m.
Right △BFT (F is the foot of the tower): h = BT sin 47° ≈ 74.9 m.

The tower is about 74.9 m tall.

Two triangles share side BT. Solve the oblique triangle ABT first, then use the right triangle BFT.

Worked example

Example 3: A four-sided lot

A cottage lot ABCD has AB = 40 m, BC = 55 m and ∠ABC = 105°. The diagonal AC splits it into two triangles. In △ACD, CD = 38 m and ∠ACD = 62°. Find the length of side AD.

△ABC (SAS): AC² = 40² + 55² − 2(40)(55) cos 105° ≈ 4625 + 1138.8 = 5763.8, so AC ≈ 75.92 m.
△ACD (SAS): AD² = 75.92² + 38² − 2(75.92)(38) cos 62° ≈ 5763.8 + 1444 − 2708.9 = 4498.9.
AD ≈ 67.1 m.

AD ≈ 67.1 m. The diagonal turned one hard problem into two cosine-law problems.

Worked example

Example 4: Posing a problem

Situation: a triangular community garden has sides of 120 m, 150 m and 200 m. Pose a question about it that needs trigonometry, and solve it.

A good question: "A gardener wants to know whether the corner opposite the 200 m side is a right angle (for a square raised bed). What is that angle?"
SSS, so use the cosine law: cos θ = (120² + 150² − 200²) ÷ (2 × 120 × 150) = −3100 ÷ 36 000 ≈ −0.0861.
θ = cos⁻¹(−0.0861) ≈ 94.9°.

The corner is about 94.9°: slightly obtuse, so a square bed would not fit snugly. (Other good questions: the smallest angle, or how far it is from one corner to the middle of the opposite side.)

Just for fun

Sailors in Canadian waters say "red right returning": when you head back into a harbour (or upstream), the red buoys stay on your right. (Much of the rest of the world uses the opposite rule.) Trigonometry tells you how far you are from port; the buoys tell you which side of them to pass on.

Hands-on activity

Measure a tall object with two sightings

You need: a protractor, a drinking straw, tape, a piece of string with a small weight (a washer or a nut), a measuring tape, and a partner. Stay on a sidewalk, path or field, away from traffic.

Make a clinometer: tape the straw along the straight edge of the protractor. Tie the string to the centre hole so the weight hangs down. When you look through the straw at something, the angle of elevation is the difference between 90° and the number where the string crosses the scale.
Pick a tall object whose base you cannot easily reach, such as a tree or a flagpole behind a fence.
Stand at point A and measure the angle of elevation to the top.
Walk straight toward the object a measured distance (say 10 m) to point B and measure the angle again.
Use the method of Example 2 to find the height above eye level, then add your eye height.

What to look for: the second angle should be bigger than the first. Results within about 10% of a known height are good; small angle errors make big differences, so measure each angle three times and average. Never look at the sun through the straw.

Your turn

You know a, b and ∠C. Which law do you use first, and why?

You know ∠A, ∠B and a. Which law?

You know ∠A, a and b. Which law, and what must you check?

A ship sails 15 km on a bearing of 070°, then 10 km on a bearing of 150°. How far is it from its starting point?

For the ship in the previous question, find the bearing of its final position from the start.

Two observers 30 m apart on the same side of a tree, in line with it, measure angles of elevation of 28° and 41° to its top. How tall is the tree (ignore eye height)?

More practice: warm-up

Write N 30° E as a true bearing.

Write the true bearing 200° in the N/S/E/W form.

What is the back bearing of 070°?

Two sides and the contained angle are known. Which law?

What is an angle of elevation measured from?

More practice: on your own

Two planes leave an airport. One flies on a bearing of 040° for 300 km, the other on a bearing of 165° for 250 km. How far apart are they?

A hiker walks 3.5 km on a bearing of 300°, then 2.8 km on a bearing of 010°. How far is she from her start?

In quadrilateral ABCD, AB = 12 cm, AD = 9 cm, ∠BAD = 78°, BC = 10 cm and CD = 8 cm. Find diagonal BD and ∠BCD.

From a boat at P, a lighthouse L is on a bearing of 040°. The boat sails 6 km due east to Q. From Q, the lighthouse is on a bearing of 330°. How far is Q from the lighthouse?

Pose your own problem about a ladder, a ramp or a roof that needs the cosine law, and solve it.

Challenge

A 30 m tower stands at the edge of a cliff. From a point A on the level ground below, the angle of elevation to the top of the tower is 52° and to the bottom of the tower is 40°. How high is the cliff?

From a boat at A, buoy B is 4 km away on a bearing of 050°. Buoy C is 3 km from B and lies on a bearing of 080° from A. How far could C be from A?

Pose a problem (with numbers) whose answer has two possible values because of the ambiguous case. Show both solutions.

Check your understanding

Quiz

Which tool solves a triangle when you know all three sides?

What is the true bearing for the direction S 20° W?

A ship's first leg has bearing 050°. At the turning point, what is the bearing back toward the start?

In Example 2, why is ∠ABT = 133°?

A ferry sails 12 km on 050°, then 9 km on 160°. What is the angle inside the triangle at the turning point?

When should you round in a multi-step problem?

For parents and teachers

Big idea (D1.6): real problems need a sketch, a plan and a choice of tool. Students combine right-triangle trigonometry, the sine law (with the ambiguous case) and the cosine law, use true bearings, and pose their own problems.

How to help: Ask your teen to explain their sketch before they calculate: "Which triangle are you solving first? What do you know about it?" For the clinometer activity, help choose a safe spot away from traffic and remind them never to sight toward the sun.

Common difficulties:

No sketch, or north lines drawn only at the start point.
Rounding a middle answer (such as BT) to one decimal place and then using it, which shifts the final answer.
Answering a different question than the one asked (for example, giving a distance when a bearing was asked).