Sine & Cosine Rule.
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Lesson Objectives
- Understand and apply the Sine Rule to find unknown sides and angles in non-right-angled triangles.
- Understand and apply the Cosine Rule to find unknown sides and angles in non-right-angled triangles.
- Interpret and solve problems involving bearings in navigation and direction-finding.
- Apply angles of elevation and depression to solve real-world height and distance problems.
- Combine multiple trigonometric tools to solve complex problems in two and three dimensions.
Introduction to Advanced Trigonometry
In Grade 10, we extend our trigonometry skills beyond right-angled triangles to solve problems involving any triangle. The Sine Rule and Cosine Rule are powerful tools for finding unknown sides and angles in non-right-angled triangles. These concepts are essential in fields like navigation, surveying, engineering, and physics. By the end of this lesson, you will be able to solve a wide variety of practical trigonometric problems involving distances, heights, and directions that cannot be solved using basic right-angle trigonometry alone.
Key Formulas for This Lesson
Sine Rule: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$
Cosine Rule (for sides): $a^2 = b^2 + c^2 - 2bc\cos A$
Cosine Rule (for angles): $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$
Area of a Triangle: $\text{Area} = \frac{1}{2}ab\sin C$
Sine Rule: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$
Cosine Rule (for sides): $a^2 = b^2 + c^2 - 2bc\cos A$
Cosine Rule (for angles): $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$
Area of a Triangle: $\text{Area} = \frac{1}{2}ab\sin C$
Key Definitions:
• Non-right-angled triangle: Any triangle that does not contain a $90^\circ$ angle.
• Included angle: The angle between two known sides in a triangle.
• Bearing: A direction measured clockwise from North, expressed as a three-digit angle (e.g., $045^\circ$).
• Angle of Elevation: The angle formed from the horizontal line of sight upwards to an object.
• Angle of Depression: The angle formed from the horizontal line of sight downwards to an object.
• Ambiguous Case (SSA): A situation where two sides and a non-included angle are known, potentially giving two possible triangles.
• Non-right-angled triangle: Any triangle that does not contain a $90^\circ$ angle.
• Included angle: The angle between two known sides in a triangle.
• Bearing: A direction measured clockwise from North, expressed as a three-digit angle (e.g., $045^\circ$).
• Angle of Elevation: The angle formed from the horizontal line of sight upwards to an object.
• Angle of Depression: The angle formed from the horizontal line of sight downwards to an object.
• Ambiguous Case (SSA): A situation where two sides and a non-included angle are known, potentially giving two possible triangles.
Quick Reference: Which Rule to Use?
| Given Information | Rule to Use | What to Find |
|---|---|---|
| Two angles and one side (AAS/ASA) | Sine Rule | Other sides |
| Two sides and a non-included angle (SSA) | Sine Rule (ambiguous) | Other angles/side |
| Two sides and included angle (SAS) | Cosine Rule | Third side |
| All three sides (SSS) | Cosine Rule | Any angle |
| Bearing problem | Draw diagram, then Sine/Cosine | Distance or bearing |
| Elevation/Depression | Trig ratios (right triangle) or Sine/Cosine | Height or distance |
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The Sine Rule
The Sine Rule relates the sides of a triangle to the sines of their opposite angles. It is used when we know:
1. Two angles and one side (AAS or ASA) — to find the other sides.
2. Two sides and a non-included angle (SSA) — to find the other angles (the ambiguous case).
$$ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} $$
Step-by-Step Method:
1. Label the triangle: side $a$ is opposite angle $A$, side $b$ opposite angle $B$, side $c$ opposite angle $C$.
2. Identify what is given and what you need to find.
3. Write the Sine Rule: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$.
4. Choose the pair of ratios that involve the given and required quantities.
5. Solve for the unknown (cross-multiply and use a calculator).
6. Check your answer (largest side opposite largest angle, etc.).
1. Label the triangle: side $a$ is opposite angle $A$, side $b$ opposite angle $B$, side $c$ opposite angle $C$.
2. Identify what is given and what you need to find.
3. Write the Sine Rule: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$.
4. Choose the pair of ratios that involve the given and required quantities.
5. Solve for the unknown (cross-multiply and use a calculator).
6. Check your answer (largest side opposite largest angle, etc.).
Triangle ABC with sides a, b, c opposite angles A, B, C respectively.
Sine Rule Example: Given A = 40°, B = 65°, a = 10 cm, find side b.
Example 1: Sine Rule (Finding a side)
Problem: In $\triangle ABC$, $A = 40^\circ$, $B = 65^\circ$, and side $a = 10$ cm. Find side $b$.
Formula: $\frac{a}{\sin A} = \frac{b}{\sin B}$
Solution:
Step 1: Write the Sine Rule: $\frac{10}{\sin 40^\circ} = \frac{b}{\sin 65^\circ}$
Step 2: Cross-multiply: $b = \frac{10 \times \sin 65^\circ}{\sin 40^\circ}$
Step 3: Calculate: $\sin 65^\circ \approx 0.9063$, $\sin 40^\circ \approx 0.6428$
$b \approx \frac{10 \times 0.9063}{0.6428} \approx \frac{9.063}{0.6428} \approx 14.10$ cm
Answer: $b \approx 14.10$ cm.
Problem: In $\triangle ABC$, $A = 40^\circ$, $B = 65^\circ$, and side $a = 10$ cm. Find side $b$.
Formula: $\frac{a}{\sin A} = \frac{b}{\sin B}$
Solution:
Step 1: Write the Sine Rule: $\frac{10}{\sin 40^\circ} = \frac{b}{\sin 65^\circ}$
Step 2: Cross-multiply: $b = \frac{10 \times \sin 65^\circ}{\sin 40^\circ}$
Step 3: Calculate: $\sin 65^\circ \approx 0.9063$, $\sin 40^\circ \approx 0.6428$
$b \approx \frac{10 \times 0.9063}{0.6428} \approx \frac{9.063}{0.6428} \approx 14.10$ cm
Answer: $b \approx 14.10$ cm.
Sine Rule (SSA): Given p = 8, q = 12, Q = 50°, find angle P.
Example 2: Sine Rule (Finding an angle, SSA)
Problem: In $\triangle PQR$, $p = 8$ cm, $q = 12$ cm, and $Q = 50^\circ$. Find angle $P$.
Formula: $\frac{p}{\sin P} = \frac{q}{\sin Q}$
Solution:
Step 1: Write the Sine Rule: $\frac{8}{\sin P} = \frac{12}{\sin 50^\circ}$
Step 2: Rearrange: $\sin P = \frac{8 \times \sin 50^\circ}{12}$
Step 3: Calculate: $\sin 50^\circ \approx 0.7660$
$\sin P \approx \frac{8 \times 0.7660}{12} = \frac{6.128}{12} \approx 0.5107$
Step 4: $P = \sin^{-1}(0.5107) \approx 30.7^\circ$
Answer: $P \approx 30.7^\circ$.
Problem: In $\triangle PQR$, $p = 8$ cm, $q = 12$ cm, and $Q = 50^\circ$. Find angle $P$.
Formula: $\frac{p}{\sin P} = \frac{q}{\sin Q}$
Solution:
Step 1: Write the Sine Rule: $\frac{8}{\sin P} = \frac{12}{\sin 50^\circ}$
Step 2: Rearrange: $\sin P = \frac{8 \times \sin 50^\circ}{12}$
Step 3: Calculate: $\sin 50^\circ \approx 0.7660$
$\sin P \approx \frac{8 \times 0.7660}{12} = \frac{6.128}{12} \approx 0.5107$
Step 4: $P = \sin^{-1}(0.5107) \approx 30.7^\circ$
Answer: $P \approx 30.7^\circ$.
Watch Out!
The Sine Rule has an ambiguous case (SSA). When given two sides and a non-included angle, there may be two possible triangles because $\sin\theta = \sin(180^\circ - \theta)$. Always check if the angle could be acute or obtuse, and draw a diagram to verify.
The Sine Rule has an ambiguous case (SSA). When given two sides and a non-included angle, there may be two possible triangles because $\sin\theta = \sin(180^\circ - \theta)$. Always check if the angle could be acute or obtuse, and draw a diagram to verify.
Practice for The Sine Rule
- In $\triangle ABC$, $A = 35^\circ$, $B = 75^\circ$, and $a = 8$ cm. Find side $b$.
- In $\triangle XYZ$, $x = 10$ cm, $y = 14$ cm, and $Y = 40^\circ$. Find angle $X$.
- In $\triangle ABC$, $A = 52^\circ$, $C = 68^\circ$, and $b = 15$ cm. Find side $a$.
- In $\triangle PQR$, $p = 6$ cm, $q = 9$ cm, and $P = 35^\circ$. Find angle $Q$.
- Explain why the SSA case can give two possible triangles, using the Sine Rule.
The Cosine Rule
The Cosine Rule is used when we know:
1. Two sides and the included angle (SAS) — to find the third side.
2. All three sides (SSS) — to find any angle.
$$ a^2 = b^2 + c^2 - 2bc\cos A $$
Step-by-Step Method:
1. Label the triangle: side $a$ opposite angle $A$, etc.
2. Identify the given information (SAS or SSS).
3. For SAS, use $a^2 = b^2 + c^2 - 2bc\cos A$ (where $a$ is the side opposite the included angle $A$).
4. For SSS, use $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ (rearrange for the other angles).
5. Solve using a calculator and check the answer (largest angle opposite largest side).
1. Label the triangle: side $a$ opposite angle $A$, etc.
2. Identify the given information (SAS or SSS).
3. For SAS, use $a^2 = b^2 + c^2 - 2bc\cos A$ (where $a$ is the side opposite the included angle $A$).
4. For SSS, use $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ (rearrange for the other angles).
5. Solve using a calculator and check the answer (largest angle opposite largest side).
Cosine Rule SAS: Given XY = 7, YZ = 9, angle Y = 60°, find XZ.
Example 3: Cosine Rule (Finding a side, SAS)
Problem: In $\triangle XYZ$, $XY = 7$ cm, $YZ = 9$ cm, and $\angle Y = 60^\circ$. Find $XZ$.
Formula: $XZ^2 = XY^2 + YZ^2 - 2(XY)(YZ)\cos Y$
Solution:
Step 1: Substitute: $XZ^2 = 7^2 + 9^2 - 2(7)(9)\cos 60^\circ$
Step 2: Calculate: $XZ^2 = 49 + 81 - 126 \times 0.5 = 130 - 63 = 67$
Step 3: $XZ = \sqrt{67} \approx 8.19$ cm
Answer: $XZ \approx 8.19$ cm.
Problem: In $\triangle XYZ$, $XY = 7$ cm, $YZ = 9$ cm, and $\angle Y = 60^\circ$. Find $XZ$.
Formula: $XZ^2 = XY^2 + YZ^2 - 2(XY)(YZ)\cos Y$
Solution:
Step 1: Substitute: $XZ^2 = 7^2 + 9^2 - 2(7)(9)\cos 60^\circ$
Step 2: Calculate: $XZ^2 = 49 + 81 - 126 \times 0.5 = 130 - 63 = 67$
Step 3: $XZ = \sqrt{67} \approx 8.19$ cm
Answer: $XZ \approx 8.19$ cm.
Cosine Rule SSS: Given a = 5, b = 6, c = 7, find angle C.
Example 4: Cosine Rule (Finding an angle, SSS)
Problem: In $\triangle ABC$, $a = 5$ cm, $b = 6$ cm, $c = 7$ cm. Find angle $C$.
Formula: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$
Solution:
Step 1: Substitute: $\cos C = \frac{5^2 + 6^2 - 7^2}{2(5)(6)}$
Step 2: Calculate: $\cos C = \frac{25 + 36 - 49}{60} = \frac{12}{60} = 0.2$
Step 3: $C = \cos^{-1}(0.2) \approx 78.46^\circ$
Answer: $C \approx 78.46^\circ$.
Problem: In $\triangle ABC$, $a = 5$ cm, $b = 6$ cm, $c = 7$ cm. Find angle $C$.
Formula: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$
Solution:
Step 1: Substitute: $\cos C = \frac{5^2 + 6^2 - 7^2}{2(5)(6)}$
Step 2: Calculate: $\cos C = \frac{25 + 36 - 49}{60} = \frac{12}{60} = 0.2$
Step 3: $C = \cos^{-1}(0.2) \approx 78.46^\circ$
Answer: $C \approx 78.46^\circ$.
Watch Out!
The Cosine Rule is more reliable than the Sine Rule for finding angles because it doesn't have an ambiguous case. However, be careful with the order of sides when using the formula — side $a$ must always be opposite angle $A$.
The Cosine Rule is more reliable than the Sine Rule for finding angles because it doesn't have an ambiguous case. However, be careful with the order of sides when using the formula — side $a$ must always be opposite angle $A$.
Practice for The Cosine Rule
- In $\triangle PQR$, $PQ = 6$ cm, $QR = 8$ cm, and $\angle Q = 45^\circ$. Find PR.
- In $\triangle ABC$, $a = 9$ cm, $b = 11$ cm, $c = 15$ cm. Find angle $A$.
- In $\triangle XYZ$, $XY = 5$ cm, $YZ = 7$ cm, and $\angle Y = 110^\circ$. Find XZ.
- In $\triangle ABC$, $a = 8$ cm, $b = 10$ cm, $c = 12$ cm. Find angle $B$.
- Explain why the Cosine Rule has no ambiguous case.
Bearings
A bearing is a direction measured clockwise from North, expressed as a three-digit angle (e.g., $045^\circ$ for Northeast, $270^\circ$ for West). Bearings are used in navigation and surveying to describe the direction from one point to another. Bearing problems typically involve forming a triangle and then applying the Sine or Cosine Rule. $$ \text{Bearing} = \text{Angle measured clockwise from North (3 digits)} $$
Step-by-Step Method:
1. Draw a diagram with North lines at each point.
2. Use the bearing to find angles between lines (alternate angles, co-interior angles).
3. Form a triangle using the given distances and angles.
4. Apply the Sine Rule or Cosine Rule to find unknown distances or bearings.
5. Convert the result back to a bearing (clockwise from North, three digits).
1. Draw a diagram with North lines at each point.
2. Use the bearing to find angles between lines (alternate angles, co-interior angles).
3. Form a triangle using the given distances and angles.
4. Apply the Sine Rule or Cosine Rule to find unknown distances or bearings.
5. Convert the result back to a bearing (clockwise from North, three digits).
Bearing Problem: Ship sails from P on bearing 060° for 20 km to Q, then on bearing 150° for 30 km to R. Find PR and bearing of R from P.
Example 5: Bearing Problem
Problem: A ship sails from port P on a bearing of $060^\circ$ for 20 km to point Q. Then it sails on a bearing of $150^\circ$ for 30 km to point R. Find the distance PR and the bearing of R from P.
Formula: Cosine Rule: $PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)\cos \angle PQR$
Solution:
Step 1: Draw the diagram. At Q, the angle between the North line and the bearing to R is $150^\circ$. The angle between the line PQ and the North line is $60^\circ$ (bearing of Q from P).
Step 2: The angle between PQ and QR at Q: $150^\circ - 60^\circ = 90^\circ$ (since the North lines are parallel). So $\angle PQR = 90^\circ$.
Step 3: Triangle PQR has sides PQ = 20 km, QR = 30 km, and included angle $90^\circ$. Use Cosine Rule to find PR:
$PR^2 = 20^2 + 30^2 - 2(20)(30)\cos 90^\circ = 400 + 900 - 0 = 1300$
$PR = \sqrt{1300} \approx 36.06$ km.
Step 4: Find $\angle QPR$ using Cosine Rule: $\cos QPR = \frac{PQ^2 + PR^2 - QR^2}{2(PQ)(PR)}$
$\cos QPR = \frac{20^2 + 1300 - 30^2}{2(20)(36.06)} = \frac{400 + 1300 - 900}{1442.4} = \frac{800}{1442.4} \approx 0.5546$
$QPR \approx \cos^{-1}(0.5546) \approx 56.3^\circ$.
Step 5: The bearing of R from P is $60^\circ + 56.3^\circ = 116.3^\circ \approx 116^\circ$.
Answer: PR $\approx 36.06$ km, bearing of R from P is $116^\circ$.
Problem: A ship sails from port P on a bearing of $060^\circ$ for 20 km to point Q. Then it sails on a bearing of $150^\circ$ for 30 km to point R. Find the distance PR and the bearing of R from P.
Formula: Cosine Rule: $PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)\cos \angle PQR$
Solution:
Step 1: Draw the diagram. At Q, the angle between the North line and the bearing to R is $150^\circ$. The angle between the line PQ and the North line is $60^\circ$ (bearing of Q from P).
Step 2: The angle between PQ and QR at Q: $150^\circ - 60^\circ = 90^\circ$ (since the North lines are parallel). So $\angle PQR = 90^\circ$.
Step 3: Triangle PQR has sides PQ = 20 km, QR = 30 km, and included angle $90^\circ$. Use Cosine Rule to find PR:
$PR^2 = 20^2 + 30^2 - 2(20)(30)\cos 90^\circ = 400 + 900 - 0 = 1300$
$PR = \sqrt{1300} \approx 36.06$ km.
Step 4: Find $\angle QPR$ using Cosine Rule: $\cos QPR = \frac{PQ^2 + PR^2 - QR^2}{2(PQ)(PR)}$
$\cos QPR = \frac{20^2 + 1300 - 30^2}{2(20)(36.06)} = \frac{400 + 1300 - 900}{1442.4} = \frac{800}{1442.4} \approx 0.5546$
$QPR \approx \cos^{-1}(0.5546) \approx 56.3^\circ$.
Step 5: The bearing of R from P is $60^\circ + 56.3^\circ = 116.3^\circ \approx 116^\circ$.
Answer: PR $\approx 36.06$ km, bearing of R from P is $116^\circ$.
Watch Out!
Bearings are always measured clockwise from North, not from the direction you are travelling. Always draw a clear diagram with North lines at each point to avoid confusion. A common error is using the bearing as an interior angle directly — you must first find the interior angle from the bearing using geometry.
Bearings are always measured clockwise from North, not from the direction you are travelling. Always draw a clear diagram with North lines at each point to avoid confusion. A common error is using the bearing as an interior angle directly — you must first find the interior angle from the bearing using geometry.
Practice for Bearings
- A ship sails from port A on a bearing of $120^\circ$ for 40 km to port B. Then it sails on a bearing of $210^\circ$ for 50 km to port C. Find the distance AC.
- An aircraft flies 100 km on a bearing of $040^\circ$, then 150 km on a bearing of $130^\circ$. Find its distance and bearing from the starting point.
- Two towns P and Q are 50 km apart. Town R is 30 km from P on a bearing of $030^\circ$ and 40 km from Q on a bearing of $120^\circ$. Find the distance PQ.
- A hiker walks 8 km on a bearing of $060^\circ$, then 6 km on a bearing of $150^\circ$. Find the distance and bearing back to the starting point.
- Explain why bearings are always written as three-digit angles.
Angles of Elevation and Depression
The angle of elevation is the angle between the horizontal line of sight and the line of sight to an object above the horizontal. The angle of depression is the angle between the horizontal line of sight and the line of sight to an object below the horizontal. These are commonly used in height and distance problems. The key relationship is that the angle of depression from a point equals the angle of elevation from the ground (alternate angles).
Step-by-Step Method:
1. Draw a diagram showing the horizontal line and the vertical height.
2. Identify the right-angled triangle (or non-right-angled triangle if there are multiple objects).
3. Apply trigonometric ratios (SOH CAH TOA) or the Sine/Cosine Rule as needed.
4. Solve for the required distance or height.
5. Check that your answer is reasonable.
1. Draw a diagram showing the horizontal line and the vertical height.
2. Identify the right-angled triangle (or non-right-angled triangle if there are multiple objects).
3. Apply trigonometric ratios (SOH CAH TOA) or the Sine/Cosine Rule as needed.
4. Solve for the required distance or height.
5. Check that your answer is reasonable.
Angle of Elevation: From point P, 50 m from the base, the angle of elevation to the top of the tower is 35°.
Example 6: Angle of Elevation
Problem: From a point 50 m away from the base of a tower, the angle of elevation to the top of the tower is $35^\circ$. Find the height of the tower.
Formula: $\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$
Solution:
Step 1: Diagram: Right-angled triangle with adjacent = 50 m, opposite = height $h$, angle = $35^\circ$.
Step 2: $\tan 35^\circ = \frac{h}{50}$
Step 3: $h = 50 \times \tan 35^\circ \approx 50 \times 0.7002 = 35.01$ m
Answer: The tower is approximately 35.0 m tall.
Problem: From a point 50 m away from the base of a tower, the angle of elevation to the top of the tower is $35^\circ$. Find the height of the tower.
Formula: $\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$
Solution:
Step 1: Diagram: Right-angled triangle with adjacent = 50 m, opposite = height $h$, angle = $35^\circ$.
Step 2: $\tan 35^\circ = \frac{h}{50}$
Step 3: $h = 50 \times \tan 35^\circ \approx 50 \times 0.7002 = 35.01$ m
Answer: The tower is approximately 35.0 m tall.
Angle of Depression: From the top of a cliff 80 m high, the angle of depression to a boat is 25°.
Example 7: Angle of Depression
Problem: From the top of a cliff 80 m high, the angle of depression to a boat at sea is $25^\circ$. Find the distance from the base of the cliff to the boat.
Formula: $\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$
Solution:
Step 1: The angle of depression from the top equals the angle of elevation from the boat (alternate angles). So the angle at the boat is $25^\circ$.
Step 2: Triangle: opposite = 80 m (height), adjacent = distance $d$, angle = $25^\circ$.
Step 3: $\tan 25^\circ = \frac{80}{d}$
Step 4: $d = \frac{80}{\tan 25^\circ} \approx \frac{80}{0.4663} \approx 171.6$ m
Answer: The boat is approximately 171.6 m from the base of the cliff.
Problem: From the top of a cliff 80 m high, the angle of depression to a boat at sea is $25^\circ$. Find the distance from the base of the cliff to the boat.
Formula: $\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$
Solution:
Step 1: The angle of depression from the top equals the angle of elevation from the boat (alternate angles). So the angle at the boat is $25^\circ$.
Step 2: Triangle: opposite = 80 m (height), adjacent = distance $d$, angle = $25^\circ$.
Step 3: $\tan 25^\circ = \frac{80}{d}$
Step 4: $d = \frac{80}{\tan 25^\circ} \approx \frac{80}{0.4663} \approx 171.6$ m
Answer: The boat is approximately 171.6 m from the base of the cliff.
Watch Out!
For angles of depression, remember that the angle of depression from the top equals the angle of elevation from the bottom because they are alternate interior angles formed by a horizontal line and a transversal. Also, ensure your calculator is in degree mode.
For angles of depression, remember that the angle of depression from the top equals the angle of elevation from the bottom because they are alternate interior angles formed by a horizontal line and a transversal. Also, ensure your calculator is in degree mode.
Practice for Angles of Elevation and Depression
- From the top of a lighthouse 70 m high, the angle of depression to a boat is $15^\circ$. Find the distance from the boat to the base of the lighthouse.
- A man standing 40 m from a tree observes the angle of elevation to the top of the tree as $28^\circ$. Find the height of the tree.
- From the top of a building 50 m high, the angle of depression to a car on the ground is $22^\circ$. Find the distance of the car from the building.
- An observer in a hot air balloon 120 m above the ground sees a landmark with an angle of depression of $35^\circ$. Find the horizontal distance from the balloon to the landmark.
- Explain why the angle of depression from a point equals the angle of elevation from the ground below it.
Applications: Navigation & Surveying
In real life, we often combine the Sine Rule, Cosine Rule, bearings, and angles of elevation/depression to solve complex problems. Surveyors use these techniques to measure distances that cannot be measured directly, such as the width of a river or the height of a mountain. These combined problems appear frequently in SSCE examinations.
Step-by-Step Method:
1. Read the problem carefully and identify what is given and what is required.
2. Draw a clear, labelled diagram showing all given information.
3. Identify which triangles are formed and what rules apply.
4. Use the appropriate rule (Sine, Cosine, SOH CAH TOA) for each triangle.
5. Combine results from multiple triangles to find the required value.
6. Check your answer for reasonableness.
1. Read the problem carefully and identify what is given and what is required.
2. Draw a clear, labelled diagram showing all given information.
3. Identify which triangles are formed and what rules apply.
4. Use the appropriate rule (Sine, Cosine, SOH CAH TOA) for each triangle.
5. Combine results from multiple triangles to find the required value.
6. Check your answer for reasonableness.
Combined Application: From the top of a building 60 m high, angles of depression to points A and B are 30° and 20° respectively.
Example 8: Combined Application
Problem: From the top of a building 60 m high, the angle of depression to point A is $30^\circ$, and the angle of depression to point B (further away) is $20^\circ$. Find the distance AB.
Formula: $\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$
Solution:
Step 1: Draw a diagram. Horizontal distance to A: $\tan 30^\circ = \frac{60}{d_A} \Rightarrow d_A = \frac{60}{\tan 30^\circ} = \frac{60}{0.5774} \approx 103.92$ m.
Step 2: Horizontal distance to B: $\tan 20^\circ = \frac{60}{d_B} \Rightarrow d_B = \frac{60}{\tan 20^\circ} = \frac{60}{0.3640} \approx 164.84$ m.
Step 3: Distance AB = $d_B - d_A = 164.84 - 103.92 = 60.92$ m.
Answer: AB $\approx 60.9$ m.
Problem: From the top of a building 60 m high, the angle of depression to point A is $30^\circ$, and the angle of depression to point B (further away) is $20^\circ$. Find the distance AB.
Formula: $\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$
Solution:
Step 1: Draw a diagram. Horizontal distance to A: $\tan 30^\circ = \frac{60}{d_A} \Rightarrow d_A = \frac{60}{\tan 30^\circ} = \frac{60}{0.5774} \approx 103.92$ m.
Step 2: Horizontal distance to B: $\tan 20^\circ = \frac{60}{d_B} \Rightarrow d_B = \frac{60}{\tan 20^\circ} = \frac{60}{0.3640} \approx 164.84$ m.
Step 3: Distance AB = $d_B - d_A = 164.84 - 103.92 = 60.92$ m.
Answer: AB $\approx 60.9$ m.
Watch Out!
In combined problems, carefully identify which triangles are right-angled and which are not. Use SOH CAH TOA only for right-angled triangles, and the Sine/Cosine Rules for non-right-angled triangles. Also, ensure you're using the correct angle of elevation/depression — the angle of depression from the top equals the angle of elevation from the bottom.
In combined problems, carefully identify which triangles are right-angled and which are not. Use SOH CAH TOA only for right-angled triangles, and the Sine/Cosine Rules for non-right-angled triangles. Also, ensure you're using the correct angle of elevation/depression — the angle of depression from the top equals the angle of elevation from the bottom.
Practice for Applications: Navigation & Surveying
- Two points P and Q are on opposite sides of a river. From point R on the bank, the angle of elevation to the top of a tree at P is $30^\circ$, and the angle of elevation to the top of the tree at Q is $45^\circ$. If the tree is 20 m tall, find the distance PQ.
- From the top of a cliff 100 m high, the angle of depression to a ship is $18^\circ$. After the ship moves 50 m directly towards the cliff, the new angle of depression is $25^\circ$. Find the original distance of the ship from the cliff.
- An observer at point A sees a tower with an angle of elevation of $15^\circ$. After walking 200 m directly towards the tower, the angle of elevation becomes $35^\circ$. Find the height of the tower.
- A surveyor measures the angle of elevation to the top of a mountain as $12^\circ$ from one point and $18^\circ$ from a point 500 m closer to the mountain. Find the height of the mountain.
- From two points on the same side of a building, the angles of elevation to the top are $24^\circ$ and $40^\circ$. If the two points are 80 m apart, find the height of the building.
Cumulative Practice Exercises
- In $\triangle ABC$, $A = 35^\circ$, $B = 75^\circ$, and $a = 8$ cm. Find side $b$.
- In $\triangle XYZ$, $x = 10$ cm, $y = 14$ cm, and $Y = 40^\circ$. Find angle $X$.
- In $\triangle PQR$, $PQ = 6$ cm, $QR = 8$ cm, and $\angle Q = 45^\circ$. Find PR.
- In $\triangle ABC$, $a = 5$ cm, $b = 6$ cm, $c = 7$ cm. Find angle $B$.
- A ship sails from port A on a bearing of $120^\circ$ for 40 km to port B. Then it sails on a bearing of $210^\circ$ for 50 km to port C. Find the distance AC.
- From the top of a lighthouse 70 m high, the angle of depression to a boat is $15^\circ$. Find the distance from the boat to the base of the lighthouse.
- Two points P and Q are on opposite sides of a river. From point R on the bank, the angle of elevation to the top of a tree at P is $30^\circ$, and the angle of elevation to the top of the tree at Q is $45^\circ$. If the tree is 20 m tall, find the distance PQ.
- In $\triangle ABC$, $A = 50^\circ$, $b = 9$ cm, $c = 11$ cm. Find side $a$.
- A plane flies 100 km on a bearing of $040^\circ$, then 150 km on a bearing of $130^\circ$. Find its distance and bearing from the starting point.
- From a point, the angle of elevation to the top of a tower is $25^\circ$. After walking 100 m towards the tower, the angle of elevation becomes $40^\circ$. Find the height of the tower.
Solutions to Cumulative Exercises
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Step 1: Sine Rule: $\frac{a}{\sin A} = \frac{b}{\sin B}$
Step 2: $b = \frac{8 \times \sin 75^\circ}{\sin 35^\circ} \approx \frac{8 \times 0.9659}{0.5736} \approx \frac{7.7272}{0.5736} \approx 13.47$ cm
Answer: $b \approx 13.47$ cm -
Step 1: Sine Rule: $\frac{x}{\sin X} = \frac{y}{\sin Y}$
Step 2: $\sin X = \frac{10 \times \sin 40^\circ}{14} \approx \frac{10 \times 0.6428}{14} \approx 0.4591$
Step 3: $X \approx \sin^{-1}(0.4591) \approx 27.3^\circ$
Answer: $X \approx 27.3^\circ$ -
Step 1: Cosine Rule: $PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)\cos Q$
Step 2: $PR^2 = 6^2 + 8^2 - 2(6)(8)\cos 45^\circ = 36 + 64 - 96 \times 0.7071 = 100 - 67.88 = 32.12$
Step 3: $PR \approx \sqrt{32.12} \approx 5.67$ cm
Answer: $PR \approx 5.67$ cm -
Step 1: $\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{5^2 + 7^2 - 6^2}{2(5)(7)} = \frac{25 + 49 - 36}{70} = \frac{38}{70} = 0.5429$
Step 2: $B = \cos^{-1}(0.5429) \approx 57.1^\circ$
Answer: $B \approx 57.1^\circ$ -
Step 1: Draw diagram. $\angle ABC = 120^\circ - 30^\circ = 90^\circ$ (since bearing of B from A is 120°, bearing of C from B is 210°, difference = 90°).
Step 2: $AC^2 = 40^2 + 50^2 - 2(40)(50)\cos 90^\circ = 1600 + 2500 = 4100$
Step 3: $AC \approx \sqrt{4100} \approx 64.03$ km
Answer: $AC \approx 64.0$ km -
Step 1: $\tan 15^\circ = \frac{70}{d} \Rightarrow d = \frac{70}{\tan 15^\circ} \approx \frac{70}{0.2679} \approx 261.3$ m
Answer: 261.3 m -
Step 1: Distance to P from R: $\tan 30^\circ = \frac{20}{d_P} \Rightarrow d_P = \frac{20}{\tan 30^\circ} = 34.64$ m
Step 2: Distance to Q from R: $\tan 45^\circ = \frac{20}{d_Q} \Rightarrow d_Q = \frac{20}{1} = 20$ m
Step 3: $PQ = d_P + d_Q = 34.64 + 20 = 54.64$ m
Answer: $PQ \approx 54.6$ m -
Step 1: $a^2 = b^2 + c^2 - 2bc\cos A = 9^2 + 11^2 - 2(9)(11)\cos 50^\circ = 81 + 121 - 198 \times 0.6428 = 202 - 127.27 = 74.73$
Step 2: $a \approx \sqrt{74.73} \approx 8.64$ cm
Answer: $a \approx 8.64$ cm -
Step 1: $\angle ABC = 130^\circ - 40^\circ = 90^\circ$ (difference in bearings).
Step 2: $AC^2 = 100^2 + 150^2 = 10000 + 22500 = 32500$
Step 3: $AC \approx \sqrt{32500} \approx 180.28$ km
Step 4: $\tan \angle BAC = \frac{150}{100} = 1.5 \Rightarrow \angle BAC \approx 56.3^\circ$
Step 5: Bearing = $040^\circ + 56.3^\circ = 96.3^\circ \approx 096^\circ$
Answer: $180.28$ km, bearing $096^\circ$ -
Step 1: Let height = $h$, initial distance = $d$. $\tan 25^\circ = \frac{h}{d} \Rightarrow h = d \tan 25^\circ$
Step 2: After walking 100 m: $\tan 40^\circ = \frac{h}{d-100} \Rightarrow h = (d-100)\tan 40^\circ$
Step 3: Equate: $d \tan 25^\circ = (d-100)\tan 40^\circ$
Step 4: $d(0.4663) = (d-100)(0.8391) \Rightarrow 0.4663d = 0.8391d - 83.91 \Rightarrow 83.91 = 0.3728d \Rightarrow d \approx 225.13$ m
Step 5: $h = 225.13 \times 0.4663 \approx 104.97$ m
Answer: Height $\approx 105.0$ m
Conclusion & Summary
In this lesson, we covered the Sine Rule and Cosine Rule for solving non-right-angled triangles, as well as bearings and angles of elevation/depression. These concepts are fundamental in navigation, surveying, and many real-world applications. By mastering these tools, you can solve a wide range of practical problems involving distances, heights, and directions that cannot be solved using basic right-angle trigonometry alone.
Key Takeaways:
1. Sine Rule: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ — use for AAS/ASA or SSA (watch the ambiguous case).
2. Cosine Rule (sides): $a^2 = b^2 + c^2 - 2bc\cos A$ — use for SAS.
3. Cosine Rule (angles): $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ — use for SSS (no ambiguous case).
4. Bearings: Measure clockwise from North, always written as three digits.
5. Angles of Elevation/Depression: Measured from the horizontal line of sight; depression from top equals elevation from bottom.
Keep practicing with different scenarios, and trigonometry will become your most versatile problem-solving tool!
Video Resource
Watch this video for more worked examples of the Sine Rule, Cosine Rule, and bearings.
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