Lesson 18 ~40 min Unit 3 · Geometry +90 XP

Mixed Trigonometry Problems

Sine rule, cosine rule, area formula. Know when to use each. Solve multi-step problems that combine all three tools in a single question.

Today's hook: A triangle has sides 8 cm and 12 cm with included angle 70°. List three different things you could find about this triangle, and which formula you would use for each.

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Think First

warm-up

A triangle has sides 8 cm and 12 cm with included angle 70°. List three different things you could find about this triangle, and which formula you would use for each. Then consider: what if you were also told the third side is 11 cm -- how would that change your approach?

Record your answer in your workbook.

The Big Idea

+5 XP to read

Every triangle problem gives you some information and asks for something else. The key skill is matching the given information to the right formula, then chaining formulas together when one is not enough.

Think of sine rule, cosine rule, and area formula as three tools in a toolkit. You do not use all three on every problem -- you pick the one that fits. Sometimes you need two: cosine rule to find a missing side, then sine rule to find an angle, then area formula to finish.

Match pattern → Choose formula → Solve

AAS/ASA → Sine

Two angles and a side.

SAS/SSS → Cosine

Two sides and included angle, or all three sides.

SAS area → Half ab sin C

Direct area from two sides and included angle.

What You'll Master

objectives

Know

When to use sine rule, cosine rule, and area formula
How to identify given and required information

Understand

Complex problems often need multiple formulas
How to verify answers using alternative methods

Can Do

Choose the correct formula for any triangle problem
Solve multi-step problems combining formulas
Check answers for reasonableness and consistency

Words You Need

vocabulary

Sine rule$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$. Use for AAS, ASA, SSA.

Cosine rule$c^2 = a^2 + b^2 - 2ab \cos C$. Use for SAS, SSS.

Area formula$A = \frac{1}{2}ab \sin C$. Use for SAS area problems.

Heron's formula$A = \sqrt{s(s-a)(s-b)(s-c)}$ where $s = \frac{a+b+c}{2}$. Use for SSS area.

Multi-stepA problem requiring two or more formulas applied in sequence.

VerifyCheck your answer using a different method or by checking reasonableness.

Spot the Trap

heads-up

Wrong: Using the sine rule for SAS problems. You cannot find the third side directly with the sine rule.

Right: SAS always requires the cosine rule for the third side, then sine rule for angles.

Wrong: Using degrees in the wrong place or mixing up angle and side labels.

Right: Always label carefully: side $a$ is opposite angle $A$. Draw a diagram first.

Which Formula to Use?

+5 XP

Use this decision tree to choose the right formula every time. The given information pattern determines your starting formula.

Given	Find	Use
AAS or ASA	Side	Sine rule
SSA	Angle or side	Sine rule (check ambiguous)
SAS	Side	Cosine rule
SAS	Area	Area formula
SSS	Angle	Cosine rule
SSS	Area	Heron's formula

Pattern → Formula → Solve

Two angles

Sine rule always works.

Two sides + angle

Is the angle included? Yes = cosine. No = sine (ambiguous).

All three sides

Cosine for angles, Heron for area.

Multi-Step Strategy

+5 XP

Many exam problems require chaining formulas together. Here is a reliable strategy that works every time.

Step 1: Draw a diagram and label all given information.
Step 2: Identify the pattern (AAS, ASA, SSA, SAS, SSS).
Step 3: Choose the first formula based on the pattern.
Step 4: Solve for the unknown.
Step 5: Use the new information to choose the next formula.
Step 6: Repeat until all required values are found.
Step 7: Verify using an alternative method or check reasonableness.

Draw → Pattern → Formula → Solve → Verify

Draw first

A diagram prevents label mix-ups.

Write the pattern

SAS, SSS, AAS -- say it aloud.

Check angles sum

A + B + C must equal 180°.

Common Pitfalls

heads-up

Using sine rule for SAS

Trying to find the third side with $\frac{a}{\sin A} = \frac{b}{\sin B}$ when you only know two sides and the included angle. The sine rule requires at least one angle-side pair, which you do not have in pure SAS.

Fix: SAS → cosine rule for the third side. Then you have enough for sine rule.

Forgetting the ambiguous case

Using $\sin^{-1}$ in SSA problems and reporting only the acute angle. When $\sin B = k$, there are two possible angles: $B$ and $180° - B$.

Fix: always check if $180° - B$ gives a valid triangle (sum of angles < 180°).

Not verifying answers

Moving on without checking. A single arithmetic error early on propagates through all subsequent calculations.

Fix: check that angles sum to 180°, or verify one value using a different formula.

Watch Me Solve It · 3 examples

Watch Me Solve It · Multi-step (SAS)

+15 XP per step

PROBLEM

Triangle with $a = 10$, $b = 12$, angle $C = 50°$. Find all sides, all angles, and the area.

1

Find side $c$ (cosine rule)

$c^2 = 10^2 + 12^2 - 2(10)(12)\cos 50°$

$c^2 = 100 + 144 - 240 \times 0.643 = 244 - 154.3 = 89.7$

$c = \sqrt{89.7} \approx 9.5$
2

Find angle $A$ (sine rule)

$\frac{\sin A}{10} = \frac{\sin 50°}{9.5}$

$\sin A = \frac{10 \times 0.766}{9.5} = 0.806$

$A = \sin^{-1}(0.806) \approx 53.7°$
3

Find angle $B$

$B = 180° - 50° - 53.7° = 76.3°$

Check: $50 + 53.7 + 76.3 = 180$ ✓
4

Find area

$A = \frac{1}{2}(10)(12)\sin 50° = 60 \times 0.766 \approx 46.0$

Could also use Heron: $s = 15.75$, $A = \sqrt{15.75 \times 5.75 \times 3.75 \times 6.25} \approx 46.0$ ✓

Answer$c \approx 9.5$, $A \approx 53.7°$, $B \approx 76.3°$, Area $\approx 46.0$

Watch Me Solve It · SSS problem

+15 XP per step

PROBLEM

Triangle ABC has $a = 7$ cm, $b = 9$ cm, $c = 11$ cm. Find all angles and the area.

1

Find angle $A$ (cosine rule)

$\cos A = \frac{9^2 + 11^2 - 7^2}{2(9)(11)} = \frac{81 + 121 - 49}{198} = \frac{153}{198} = 0.773$

$A = \cos^{-1}(0.773) \approx 39.4°$
2

Find angle $B$ (cosine rule)

$\cos B = \frac{7^2 + 11^2 - 9^2}{2(7)(11)} = \frac{49 + 121 - 81}{154} = \frac{89}{154} = 0.578$

$B = \cos^{-1}(0.578) \approx 54.7°$
3

Find angle $C$

$C = 180° - 39.4° - 54.7° = 85.9°$

Check: sum = 180° ✓
4

Find area (Heron's formula)

$s = \frac{7+9+11}{2} = 13.5$

$A = \sqrt{13.5 \times 6.5 \times 4.5 \times 2.5} = \sqrt{988.6} \approx 31.4$ cm²

Verify with area formula: $\frac{1}{2}(7)(9)\sin 85.9° \approx 31.4$ ✓

Answer$A \approx 39.4°$, $B \approx 54.7°$, $C \approx 85.9°$, Area $\approx 31.4$ cm²

Watch Me Solve It · Bearings problem

+15 XP per step

PROBLEM

Three towns A, B, C form a triangle. $AB = 25$ km, $AC = 30$ km. The bearing of B from A is 070°, and the bearing of C from A is 130°. Find $BC$, the area, and the cost of a road from B to C at $\$50{,}000$ per km.

1

Find angle $BAC$

$\angle BAC = 130° - 70° = 60°$

The difference in bearings gives the included angle.
2

Find $BC$ (cosine rule)

$BC^2 = 25^2 + 30^2 - 2(25)(30)\cos 60°$

$BC^2 = 625 + 900 - 1500 \times 0.5 = 1525 - 750 = 775$

$BC = \sqrt{775} \approx 27.8$ km
3

Find the area

$A = \frac{1}{2}(25)(30)\sin 60° = 375 \times 0.866 \approx 324.8$ km²
4

Find the cost

Cost = $27.8 \times 50{,}000 \approx \$1{,}390{,}000$

Approximately 1.4 million dollars.

Answer$BC \approx 27.8$ km, Area $\approx 324.8$ km², Cost $\approx \$1.39$ million

Copy Into Your Books

AAS / ASA

Sine rule
Find missing angle first (180 - A - B)
Then use sine rule for sides

SSA

Sine rule (ambiguous case)
Check if 180 - B is valid
Two possible triangles possible

SAS

Cosine rule for third side
Sine rule for angles
Area formula for area

SSS

Cosine rule for angles
Heron's formula for area
Verify: angles sum to 180°

How are you completing this lesson?

Brain Trainer · 4 problems

Brain Trainer · Mixed drill

4 problems

Four quick problems. Identify the formula first, then solve.

1 A triangle has $a = 11$ cm, $b = 14$ cm, angle $C = 65°$. Find side $c$, angle $A$, and the area.

$c^2 = 121 + 196 - 308\cos 65° = 186.7$, $c \approx 13.7$ cm. $\sin A = \frac{11 \sin 65°}{13.7} = 0.728$, $A \approx 46.7°$. Area = $\frac{1}{2}(11)(14)\sin 65° \approx 69.8$ cm².$c \approx 13.7$ cm, $A \approx 46.7°$, Area $\approx 69.8$ cm²
2 A triangle has area 30 cm², sides $a = 10$ and $b = 8$. Find the included angle $C$.

$30 = \frac{1}{2}(10)(8)\sin C = 40\sin C$. $\sin C = 0.75$. $C = 48.6°$ or $C = 131.4°$.$C = 48.6°$ or $131.4°$
3 A triangle has sides 9, 12, 15. What type of triangle is it? Verify using the cosine rule.

$9^2 + 12^2 = 81 + 144 = 225 = 15^2$. It is a right triangle. Cosine rule: $\cos C = \frac{81+144-225}{216} = 0$, so $C = 90°$.Right triangle, $C = 90°$
4 A triangle has $a = 6$, $b = 8$, $c = 10$. Find all angles and the area using two different methods.

Recognise 6-8-10 as scaled 3-4-5 right triangle. $C = 90°$, $\sin A = 6/10 = 0.6$, $A = 36.9°$, $B = 53.1°$. Area = $\frac{1}{2}(6)(8) = 24$. Heron: $s = 12$, $A = \sqrt{12 \times 6 \times 4 \times 2} = \sqrt{576} = 24$.$A = 36.9°$, $B = 53.1°$, $C = 90°$, Area = 24

Complete in your workbook.

Multiple Choice · 5 questions

Formula selection (AAS)

+10 XP

Given AAS (two angles and a non-included side), which formula finds a missing side?

Sine rule

Cosine rule

Area formula

Pythagoras

A is correct. The sine rule $\frac{a}{\sin A} = \frac{b}{\sin B}$ is the direct method when you know angles and one side.

Formula selection (SAS)

+10 XP

Given SAS (two sides and the included angle), which formula finds the third side?

Sine rule

Cosine rule

Area formula

Heron's formula

B is correct. The cosine rule $c^2 = a^2 + b^2 - 2ab\cos C$ directly gives the third side when two sides and the included angle are known.

Ambiguous case

+10 XP

A triangle has area 24, sides 8 and 10. The included angle is:

30°

36.9°

36.9° or 143.1°

45°

C is correct. $24 = \frac{1}{2}(8)(10)\sin C = 40\sin C$, so $\sin C = 0.6$. This gives $C = 36.9°$ or $C = 180° - 36.9° = 143.1°$. Both are valid since the angle sum with other angles would still be less than 180°.

Formula selection (SSS)

+10 XP

Given SSS (all three sides), the best first step to find an angle is:

Sine rule to find an angle

Cosine rule to find an angle

Area formula

Heron's formula

B is correct. The cosine rule can be rearranged to $\cos C = \frac{a^2+b^2-c^2}{2ab}$, giving an angle directly from three sides. The sine rule needs at least one angle to start.

Triangle classification

+10 XP

A triangle with sides 9, 12, 15 is:

Acute

Right

Obtuse

Equilateral

B is correct. $9^2 + 12^2 = 81 + 144 = 225 = 15^2$. This satisfies Pythagoras' theorem, so it is a right triangle with hypotenuse 15.

Short Answer · 3 questions

Multi-step SAS problem

+15 XP

In triangle ABC, $a = 11$ cm, $b = 14$ cm, angle $C = 65°$.

(a) Find side $c$.

(b) Find angle $A$.

(a) $c^2 = 11^2 + 14^2 - 2(11)(14)\cos 65° = 121 + 196 - 308 \times 0.423 = 317 - 130.3 = 186.7$

$c = \sqrt{186.7} \approx 13.7$ cm

(b) $\frac{\sin A}{11} = \frac{\sin 65°}{13.7}$. $\sin A = \frac{11 \times 0.906}{13.7} = 0.728$. $A \approx 46.7°$

(c) Area = $\frac{1}{2}(11)(14)\sin 65° = 77 \times 0.906 \approx 69.8$ cm²

Quadrilateral problem

+15 XP

A quadrilateral ABCD has diagonal $AC = 20$ cm. The other diagonal $BD = 16$ cm intersects $AC$ at $E$. $AE = 8$ cm, $EC = 12$ cm, and $\angle AEB = 70°$.

(a) Find the area of triangle AEB.

(b) Find $BE$ using the cosine rule in triangle AEB (given $AB = 10$ cm).

(b) $10^2 = 8^2 + BE^2 - 2(8)(BE)\cos 70°$

$100 = 64 + BE^2 - 5.47 BE$

$BE^2 - 5.47 BE - 36 = 0$

$BE = \frac{5.47 + \sqrt{29.9 + 144}}{2} = \frac{5.47 + 13.2}{2} \approx 9.3$ cm

(a) Area AEB = $\frac{1}{2}(8)(9.3)\sin 70° \approx 35.0$ cm²

(c) Area CEB = $\frac{1}{2}(12)(9.3)\sin 110° \approx 52.5$ cm². Total = $35.0 + 52.5 = 87.5$ cm²

Bearings and cost

+15 XP

Three towns A, B, C form a triangle. $AB = 25$ km, $AC = 30$ km. The bearing of B from A is 070°, and the bearing of C from A is 130°.

(a) Find angle $BAC$.

(b) Find the distance $BC$.

(d) A new road is to be built directly from B to C. If road construction costs $\$50{,}000$ per km, estimate the cost.

(a) $\angle BAC = 130° - 70° = 60°$

(b) $BC^2 = 25^2 + 30^2 - 2(25)(30)\cos 60° = 625 + 900 - 750 = 775$. $BC = \sqrt{775} \approx 27.8$ km

(c) Area = $\frac{1}{2}(25)(30)\sin 60° = 375 \times 0.866 \approx 324.8$ km²

(d) Cost = $27.8 \times 50{,}000 \approx \$1{,}390{,}000$

Stretch Challenge

+25 XP

A triangle has sides $a = 13$, $b = 14$, $c = 15$. Without using Heron's formula, find its area by first finding an angle using the cosine rule, then using the area formula. Verify your answer using Heron's formula. Which method do you prefer, and why?

Method 1 (cosine then area):

$\cos C = \frac{13^2 + 14^2 - 15^2}{2(13)(14)} = \frac{169 + 196 - 225}{364} = \frac{140}{364} = 0.385$

$C = \cos^{-1}(0.385) \approx 67.4°$

Area = $\frac{1}{2}(13)(14)\sin 67.4° = 91 \times 0.923 \approx 84.0$

Method 2 (Heron):

$s = \frac{13+14+15}{2} = 21$

$A = \sqrt{21 \times 8 \times 7 \times 6} = \sqrt{7056} = 84$

Conclusion: Heron's formula gives an exact integer (84) while the cosine method introduces rounding error. For SSS, Heron is often cleaner when the numbers work out nicely.

Quick Review

all together

AAS / ASA → Sine rule

Two angles mean sine rule is the direct path.

SAS → Cosine then sine

Cosine for the third side, sine for angles, area formula for area.

SSS → Cosine then Heron

Cosine rule for angles, Heron's formula for area.

SSA → Sine rule (ambiguous)

Check if $180° - B$ gives a valid second triangle.

Verify every answer

Angles sum to 180°. Use alternative methods to check.

Draw a diagram first

Prevents label mix-ups and reveals the pattern.

Interactive Method Chooser

+10 XP

Test your ability to choose the correct formula for each problem type. Sine rule, cosine rule, area formula, or Pythagoras?

Badge Wall

badges

First Step

On Fire

Perfectionist

Speedster

All Phases

Completionist

Daily Challenge

+15 XP

A triangle has $a = 8$ cm, $b = 10$ cm, $c = 12$ cm. Find all three angles and the area using two different methods. Show which method you prefer and explain why.

Lesson Complete

+20 XP

You have finished Lesson 18 -- Mixed Trigonometry Problems. You can now choose the right formula for any triangle problem and chain multiple formulas together to solve complex multi-step questions.

11 cards studied

4 drills completed

5 MCQs

3 SAQs

3 formulas mastered