Ch 9: Some Applications of Trigonometry

Heights and Distances — Introduction

Chapter 9: Some Applications of Trigonometry

Page 1 of 5: Heights and Distances — Introduction

Welcome to the practical world of trigonometry! In the last chapter, you mastered trigonometric ratios like sine, cosine, and tangent. Now, we'll see how these powerful tools help us measure the world around us without a measuring tape.

Imagine you want to find the height of a very tall tree, a skyscraper, or a mountain. You can't just climb it with a ruler! This is where trigonometry comes to the rescue. By measuring an angle and a distance along the ground, you can calculate incredible heights and distances. This branch of mathematics, often called Heights and Distances, is used by surveyors, engineers, astronomers, and navigators. In this lesson, we will learn the fundamental concepts that form the building blocks for solving these real-world problems.

{{KEY: type=concept | title=The Core Idea | text=By forming a right-angled triangle using our position, the object, and the ground, we can use a known side (like the distance from the object) and a known angle (the angle we look up at) to find an unknown side (like the height of the object) using trigonometric ratios.}}

Definitions & Key Terms

To solve problems involving heights and distances, we must first understand the specific terms used to describe the scenario. These terms help us translate a real-world situation into a mathematical diagram.

{{VISUAL: diagram: Two scenarios side-by-side. Left side shows an observer looking up at a kite, with a horizontal line from their eye, the line of sight to the kite, and the angle of elevation labeled θ. Right side shows an observer on a cliff looking down at a boat, with a horizontal line from their eye, the line of sight to the boat, and the angle of depression labeled α.}}

The Problem-Solving Logic

Every "Heights and Distances" problem can be solved by following a clear, logical process. The goal is always to create a right-angled triangle from the given information and then apply trigonometry.

1. Visualize and Draw: Read the problem carefully and draw a simple, neat diagram. This is the most crucial step. Represent objects like trees, towers, or buildings with vertical lines. Represent the ground with a horizontal line. Label the points (e.g., A, B, C) and mark the right angle.

2. Identify the Triangle: Locate the right-angled triangle in your diagram. Mark the given information on it – the lengths of sides you know and the angles you know.

3. Label the Sides: With respect to the known angle (other than the 90° angle), label the sides of the triangle as Opposite, Adjacent, and Hypotenuse.

   • Opposite: The side directly across from the angle.
   • Adjacent: The side next to the angle (that is not the hypotenuse).
   • Hypotenuse: The side opposite the right angle.

4. Choose the Right Ratio: Decide which trigonometric ratio connects the side you know with the side you need to find. The mnemonic SOH-CAH-TOA is your best friend here.

   • SOH: Sin θ = Opposite / Hypotenuse
   • CAH: Cos θ = Adjacent / Hypotenuse
   • TOA: Tan θ = Opposite / Adjacent

5. Formulate and Solve: Write the equation using the chosen ratio, substitute the known values, and solve for the unknown side.

6. Final Check: Reread the question to ensure you've answered what was asked. For example, if the observer's height was given, remember to add it to the calculated height from your triangle to find the total height of the object.

Solved Examples

Let's apply this logic to some problems, starting from easy and moving to more complex scenarios.

Example 1: Finding the Height of a Tower (Easy)

Given: From a point on the ground 15 m away from the foot of a vertical tower, the angle of elevation to the top of the tower is 60°.

To Find: The height of the tower.

Solution:

1. First, let's draw a diagram. Let AB be the tower and C be the point on the ground. This forms a right-angled triangle ABC, with the right angle at B.

{{VISUAL: diagram: A right-angled triangle ABC, with the right angle at B. AB is the vertical side representing the tower. BC is the horizontal side representing the ground, labeled '15 m'. The angle at C (∠ACB) is the angle of elevation, labeled '60°'. The height of the tower, AB, is labeled 'h'.}}

2. We are given the adjacent side BC = 15 m and the angle ∠ACB = 60°. We need to find the opposite side AB (the height).

3. The ratio that connects the Opposite and Adjacent sides is the tangent (tan).

tan C = Opposite / Adjacent = AB / BC

4. Substitute the known values. We know that tan 60° = √3.

tan 60° = AB / 15

√3 = AB / 15

5. Solve for AB.

AB = 15 × √3

Final Answer: The height of the tower is 15√3 m.

Example 2: The Electrician's Ladder (Medium)

Given: An electrician needs to repair a fault on a pole of height 5 m. The fault is 1.3 m below the top. The ladder must be inclined at 60° to the horizontal.

To Find: The length of the ladder and the distance of the foot of the ladder from the pole. (Use √3 ≈ 1.73)

Solution:

1. Let AD be the pole. The electrician needs to reach point B, where BD = AD - AB = 5 - 1.3 = 3.7 m. Let BC be the ladder. BDC is a right-angled triangle.

2. We have the height to reach (Opposite side BD = 3.7 m) and the angle ∠BCD = 60°. We need to find the ladder length (Hypotenuse BC) and the distance from the pole (Adjacent side DC).

3. To find the ladder length (Hypotenuse BC): We have the Opposite side and need the Hypotenuse. The sine ratio connects these.

sin 60° = Opposite / Hypotenuse = BD / BC

4. Substitute the values. We know sin 60° = √3 / 2.

√3 / 2 = 3.7 / BC

5. Solve for BC.

BC = (3.7 × 2) / √3 = 7.4 / 1.73 ≈ 4.28

6. To find the distance from the pole (Adjacent DC): We have the Opposite side BD and need the Adjacent DC. The cotangent ratio connects these.

cot 60° = Adjacent / Opposite = DC / BD

7. Substitute the values. We know cot 60° = 1/√3.

1/√3 = DC / 3.7

8. Solve for DC.

DC = 3.7 / √3 = 3.7 / 1.73 ≈ 2.14

Final Answer: The length of the ladder should be 4.28 m (approx.), and it should be placed 2.14 m (approx.) from the foot of the pole.

Example 3: The Chimney and the Observer (Hard)

Given: An observer, 1.5 m tall, is 28.5 m away from a chimney. The angle of elevation of the top of the chimney from her eyes is 45°.

To Find: The height of the chimney.

Solution:

1. Draw a diagram. Let AB be the chimney and CD be the observer. Draw a horizontal line DE from the observer's eye D to the chimney. This forms a right-angled triangle ADE.

{{VISUAL: diagram: A chimney AB and an observer CD. A horizontal dashed line DE is drawn from D to AB. The rectangle CBED is formed. CD = BE = 1.5m. CB = DE = 28.5m. The angle of elevation ∠ADE is labeled 45°. The height of the chimney AB = AE + EB.}}

2. Here, CD = BE = 1.5 m (height of observer). The distance CB = DE = 28.5 m. The angle of elevation is ∠ADE = 45°.

3. We need to find the total height of the chimney, AB = AE + BE. We already know BE = 1.5 m. So, we first need to find AE.

4. In the right-angled triangle ADE, we know the Adjacent side DE = 28.5 m and the angle ∠ADE = 45°. We need to find the Opposite side AE.

5. The tangent ratio connects Opposite and Adjacent.

tan 45° = Opposite / Adjacent = AE / DE

6. Substitute the values. We know tan 45° = 1.

1 = AE / 28.5

AE = 28.5 m

7. Now, calculate the total height of the chimney AB.

AB = AE + BE = 28.5 + 1.5 = 30

Final Answer: The height of the chimney is 30 m.

Example 4: The Flagstaff on the Building (Tricky)

Given: From a point P on the ground, the angle of elevation to the top of a 10 m tall building is 30°. A flag is hoisted on top of the building, and the angle of elevation to the top of the flagstaff from P is 45°.

To Find: The length of the flagstaff and the distance of the building from point P. (Use √3 ≈ 1.732)

Solution:

1. Let AB be the 10 m tall building and BD be the flagstaff. Let P be the point on the ground. This gives us two right-angled triangles: PAB and PAD.

2. In triangle PAB: AB = 10 m, ∠APB = 30°. We need to find PA (distance) and BD (part of the height in the second triangle).

3. Let's first find the distance PA using triangle PAB. We have the Opposite side AB and need the Adjacent side PA.

tan 30° = AB / PA

4. Substitute the values. We know tan 30° = 1/√3.

1/√3 = 10 / PA

PA = 10√3 ≈ 10 × 1.732 = 17.32 m

5. Now, let's consider the larger triangle PAD. The total height is AD = AB + BD = 10 + BD. The angle of elevation is ∠APD = 45°. The base is PA = 10√3 m.

6. Using the tan ratio for triangle PAD:

tan 45° = AD / PA

7. Substitute the known values. We know tan 45° = 1.

1 = (10 + BD) / (10√3)

8. Solve for BD.

10√3 = 10 + BD

BD = 10√3 - 10 = 10(√3 - 1)

9. Now, calculate the numerical value of BD.

BD = 10(1.732 - 1) = 10(0.732) = 7.32 m

Final Answer: The length of the flagstaff is 7.32 m and the distance of the building from point P is 17.32 m.

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

1. From the top of a 75 m high lighthouse, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

   💡 Answer: Let the lighthouse be AB = 75 m. Let the ships be at C and D. In ΔABC, tan 45° = 75/BC → BC = 75 m. In ΔABD, tan 30° = 75/BD → BD = 75√3 m. The distance between ships is CD = BD - BC = 75√3 - 75 = 75(√3 - 1) m.

2. The shadow of a tower standing on level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it is 60°. Find the height of the tower.

   💡 Answer: Let height be h. Let the shorter shadow be x. Then tan 60° = h/x → x = h/√3. The longer shadow is x+40. So, tan 30° = h/(x+40). Substitute x: 1/√3 = h/((h/√3) + 40). Solving this gives h = 20√3 m.

3. An aeroplane, when flying at a height of 3000 m from the ground, passes vertically above another aeroplane at an instant when the angles of elevation of the two planes from the same point on the ground are 60° and 45° respectively. What is the vertical distance between the aeroplanes at that instant?

   💡 Answer: Let the higher plane be at B and lower at C. Ground point is P. Height AB = 3000m. In ΔPAB, tan 60° = 3000/PA → PA = 3000/√3 = 1000√3 m. In ΔPAC, tan 45° = AC/PA → AC = PA = 1000√3 m. Vertical distance = AB - AC = 3000 - 1000√3 = 1000(3 - √3) m.

Mini Cheatsheet

Heights and Distances — Part 1 (Basic Applications)

Page 2: Heights and Distances — Part 1 (Basic Applications)

Welcome to the practical world of trigonometry! In the previous chapter, we learned about trigonometric ratios like sine, cosine, and tangent. Now, we'll see how these powerful tools help us measure the world around us without a measuring tape.

Imagine you're standing on the ground, looking up at the very top of the Qutub Minar. How could you figure out its height without climbing it? Trigonometry provides the answer. By measuring your distance from the base of the minar and the angle at which you have to tilt your head up to see the top, you can calculate its height with surprising accuracy. This is the essence of "heights and distances"—using angles and known lengths to find unknown lengths. This skill is used every day by architects, surveyors, astronomers, and engineers to build bridges, map lands, and even track celestial bodies.

{{FORMULA: expr=SOH-CAH-TOA | symbols=sin θ:Opposite/Hypotenuse, cos θ:Adjacent/Hypotenuse, tan θ:Opposite/Adjacent}}

Definitions & Key Terms

Before we start solving problems, let's master the vocabulary. These four terms are the foundation of every heights and distances problem.

{{VISUAL: diagram: Two simple figures side-by-side. The left figure shows an observer looking up at a bird, with the horizontal line, line of sight, and the Angle of Elevation clearly labeled. The right figure shows an observer on a cliff looking down at a boat, labeling the horizontal line, line of sight, and the Angle of Depression.}}

The Problem-Solving Framework

Solving these problems involves a logical, step-by-step process. The core idea is to translate a real-world scenario into a right-angled triangle, which we can then solve using trigonometry.

1. Visualize and Draw: Read the problem carefully and draw a simple, clear diagram. This is the most crucial step. Represent objects like towers, buildings, or trees as vertical lines, and the ground as a horizontal line. This will almost always form a right-angled triangle.

2. Label the Diagram: Mark all the given information on your diagram. Label the vertices of the triangle (e.g., A, B, C). Write down the known lengths and the angle of elevation or depression. Identify the side you need to find and label it with a variable (like h for height or x for distance).

3. Identify the Triangle and Angle: Focus on the right-angled triangle. Identify the angle you are working with (e.g., 30°, 45°, 60°). Relative to this angle, label the sides as Opposite (the side facing the angle), Adjacent (the side next to the angle, not the hypotenuse), and Hypotenuse (the side opposite the right angle).

4. Select the Right Trigonometric Ratio: Use the mnemonic SOH-CAH-TOA to decide which ratio to use.

   • If you know the Adjacent side and need to find the Opposite side (or vice versa), use Tan (tan θ = O/A).
   • If you know the Hypotenuse and need to find the Opposite side (or vice versa), use Sin (sin θ = O/H).
   • If you know the Hypotenuse and need to find the Adjacent side (or vice versa), use Cos (cos θ = A/H).

5. Formulate and Solve the Equation: Write the equation using the chosen ratio, the known angle, the known side, and the unknown variable. Solve the equation to find the value of the unknown.

6. Final Check: Reread the question to ensure you've answered what was asked. For example, if the observer's height was involved, did you remember to add it to the calculated height? Make sure your answer includes the correct units (e.g., meters, cm).

Solved Examples

Let's apply this framework to some problems, starting from easy and moving to more complex scenarios.

Example 1: The Classic Tower Problem (Easy)

A tower stands vertically on the ground. From a point on the ground, which is 15 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60°. Find the height of the tower.

Given: Distance from the foot of the tower = 15 m, Angle of elevation = 60°.

To Find: Height of the tower.

Solution:

1. Let's represent the tower by AB and the point on the ground by C. This forms a right-angled triangle ABC, with the right angle at B.

{{VISUAL: diagram: A right-angled triangle ABC, with the right angle at B. AB is the vertical side labeled 'h' (height of the tower). BC is the horizontal side labeled '15 m'. The angle at C is marked as 60°.}}

2. Here, BC is the Adjacent side to ∠ACB, and AB is the Opposite side. We need to find the Opposite side and we know the Adjacent side. Therefore, we should use the tan ratio.

3. Set up the equation using tan θ = Opposite / Adjacent.

tan 60° = AB / BC

4. Substitute the known values into the equation. We know that tan 60° = √3.

√3 = AB / 15

5. Solve for AB (the height of the tower).

AB = 15√3

Final Answer: The height of the tower is 15√3 m.

Example 2: The Chimney and the Observer (Medium)

An observer 1.5 m tall is 28.5 m away from a chimney. The angle of elevation of the top of the chimney from her eyes is 45°. What is the height of the chimney?

Given: Observer's height = 1.5 m, Distance from chimney = 28.5 m, Angle of elevation = 45°.

To Find: Total height of the chimney.

Solution:

1. Let AB be the chimney and CD be the observer. Draw a line DE parallel to the ground CB. The right-angled triangle is ADE.

{{VISUAL: diagram: A figure showing a vertical chimney AB and a shorter vertical line CD for the observer. A horizontal dashed line DE is drawn from the observer's eye D to the chimney. The distance CB is labeled 28.5 m. The observer's height CD is 1.5 m. The angle of elevation, ∠ADE, is marked as 45°. The part of the chimney AE is labeled 'h'.}}

2. We need to find the total height AB, which is AE + EB. We know EB = CD = 1.5 m. The distance DE = CB = 28.5 m. We need to find AE.

3. In the right-angled triangle ADE, AE is the Opposite side to the 45° angle, and DE is the Adjacent side. We use the tan ratio.

tan 45° = AE / DE

4. Substitute the known values. We know that tan 45° = 1.

1 = AE / 28.5

5. Solve for AE.

AE = 28.5 m

6. Now, find the total height of the chimney AB.

AB = AE + EB = 28.5 + 1.5

AB = 30 m

Final Answer: The height of the chimney is 30 m.

Example 3: The Broken Tree (Hard)

A tree breaks due to a storm and the broken part bends so that the top of the tree touches the ground, making an angle of 30° with it. The distance from the foot of the tree to the point where the top touches the ground is 8 m. Find the original height of the tree.

Given: Distance from foot to top on ground = 8 m, Angle with the ground = 30°.

To Find: Original height of the tree.

Solution:

1. Let the tree be AC. It breaks at point B. The broken part BA touches the ground at point D. So, AD is the broken part, and its length is equal to AB. The remaining part of the tree is BC. The original height is AB + BC.

2. We have a right-angled triangle BCD, with the right angle at C. CD = 8 m and ∠BDC = 30°. We need to find BC and BD (since BD is the broken part, its length is AB).

3. First, let's find BC (Opposite side) using the tan ratio with CD (Adjacent side).

tan 30° = BC / CD

1/√3 = BC / 8

BC = 8/√3 m

4. Next, let's find BD (Hypotenuse) using the cos ratio with CD (Adjacent side).

cos 30° = CD / BD

√3/2 = 8 / BD

BD = (8 × 2) / √3 = 16/√3 m

5. The original height of the tree was BC + AB. Since AB is the broken part which is now BD, the original height is BC + BD.

Height = (8/√3) + (16/√3)

Height = 24/√3 m

6. To rationalize the denominator, multiply the numerator and denominator by √3.

Height = (24 × √3) / (√3 × √3) = 24√3 / 3

Height = 8√3 m

Final Answer: The original height of the tree was 8√3 m.

Example 4: The View from the Cliff (Tricky)

From the top of a 75 m high lighthouse from the sea level, the angle of depression of a ship is 30°. Find the distance of the ship from the lighthouse.

Given: Height of lighthouse = 75 m, Angle of depression = 30°.

To Find: Distance of the ship from the lighthouse.

Solution:

1. Let AB be the lighthouse and C be the position of the ship. Let X be a point on the horizontal line from A. The angle of depression is ∠XAC = 30°.

2. The line AX is parallel to the ground line BC. Therefore, the angle of elevation from the ship, ∠ACB, is equal to the angle of depression, ∠XAC, because they are alternate interior angles. So, ∠ACB = 30°.

3. We now have a right-angled triangle ABC, with AB = 75 m, ∠ACB = 30°, and we need to find BC.

4. In ΔABC, AB is the Opposite side to the 30° angle, and BC is the Adjacent side. We should use the tan ratio.

tan 30° = AB / BC

5. Substitute the known values. We know tan 30° = 1/√3.

1/√3 = 75 / BC

6. Solve for BC.

BC = 75√3 m

Final Answer: The distance of the ship from the lighthouse is 75√3 m.

{{KEY: type=concept | title=Angle of Depression & Alternate Interior Angles | text=A very common trick in problems is using the angle of depression. Remember that the angle of depression from the top of an object is always equal to the angle of elevation from the bottom object, because they form a 'Z' shape with parallel lines (the horizontal line and the ground).}}

Tips & Tricks

Use these shortcuts to solve problems faster and more confidently.

Common Mistakes to Avoid

Many students make the same few mistakes. Watch out for these!

Brain-Teaser Questions

1. You are standing 50 m away from a tall building. As you walk 30 m closer to the building, does the angle of elevation to its top double? Why or why not?

   💡 Answer: No, it does not double. The angle of elevation is given by tan θ = Height / Distance. The tan function is not a linear function. As the distance decreases, tan θ increases, but not proportionally. For the angle to double, tan(2θ) would have to equal Height / (New Distance), which is not generally true.

2. A ladder is leaning against a wall, making an angle of 60° with the ground. If the foot of the ladder is pulled away from the wall by 2 m, the angle changes to 30°. How high up the wall does the ladder reach initially?

   💡 Answer: This requires setting up two equations. Let the initial distance from the wall be x and the height be h. Eq 1: tan 60° = h/x → h = x√3. Eq 2: tan 30° = h/(x+2) → h = (x+2)/√3. Equating both expressions for h: x√3 = (x+2)/√3 → 3x = x+2 → 2x = 2 → x=1. The initial height is h = x√3 = 1√3 = √3 m.

3. Two poles of equal height are standing opposite each other on either side of a road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. What is the position of the point?

   💡 Answer: Let the height of the poles be h. Let the point be x meters from the first pole. Its distance from the second pole is 80-x. Eq 1: tan 60° = h/x → h = x√3. Eq 2: tan 30° = h/(80-x) → h = (80-x)/√3. Equating h: x√3 = (80-x)/√3 → 3x = 80-x → 4x = 80 → x = 20. The point is 20 m from the first pole (the one with the 60° angle) and 60 m from the second pole.

Mini Cheatsheet

Screenshot this table for quick revision before your exam!

Heights and Distances — Part 2 (Multiple Objects/Observations)

Heights and Distances — Part 2 (Multiple Objects/Observations)

Welcome back! In the previous section, we tackled problems involving a single right-angled triangle. Now, we'll level up. Real-world scenarios often involve more complexity: observing an object from two different points, or looking at two different objects from the same point.

These situations create two (or more) right-angled triangles that are linked together. The key to solving them is finding the "bridge"—a common side or a shared relationship—that connects the triangles. By setting up a trigonometric equation for each triangle, you can solve them simultaneously to find the missing heights or distances.

{{FORMULA: expr=tan θ = Opposite / Adjacent | symbols=θ:angle of elevation/depression, Opposite:height of object, Adjacent:distance from object}}

Concept Introduction

Imagine you're flying a drone to capture footage of a tall monument. First, you position the drone some distance away and tilt its camera up at a 30° angle to get the whole monument in the frame. This forms one right-angled triangle.

Then, to get a more dramatic shot, you fly the drone 100 meters closer to the monument. Now, you have to tilt the camera up more steeply, say at a 60° angle. This forms a second right-angled triangle. The monument's height is the same in both scenarios—it's the common side.

Using the information from both observations (the two angles and the distance you moved), you can calculate not only the monument's height but also your initial distance from it, all without a single measuring tape. This is the power of using multiple triangles in trigonometry.

The Logic: Linking Two Triangles

The core strategy for solving problems with two angles of observation is to create a system of two equations with two variables and solve them.

1. Draw and Label: Sketch a clear diagram representing the situation. Label all known points, distances, and angles. Identify the two right-angled triangles in your sketch.
2. Identify the Link: Find the side that is common to both triangles (like the height of a tower) or a side that is part of a known total length (like the distance between two observation points). This "link" is crucial.
3. Form Equation 1: Write a trigonometric equation for the first triangle using one of the unknown variables (e.g., height h or distance x). The most common ratio to use is tan θ.
4. Form Equation 2: Write a second trigonometric equation for the second triangle, using the same variables.
5. Solve Simultaneously: You now have two equations. Use substitution or elimination to solve for the unknown variables. Often, you'll solve for one variable in terms of the other in one equation and substitute it into the second.
6. Calculate the Final Answer: Substitute the value of the first variable you found back into an earlier equation to find the second variable. Ensure your answer includes the correct units (e.g., meters).

{{VISUAL: diagram: Two right-angled triangles, ΔABC and ΔDBC, sharing a common side BC (height 'h'). Point A and D are on the ground. Angle of elevation from A is α, and from D is β. The distance AD is marked as 'd'.}}

{{KEY: type=concept | title=The Common Side Strategy | text=In problems with two triangles, the solution almost always involves expressing a common side (like the height of a tower) in terms of other variables from both triangles, and then equating the two expressions to solve for an unknown distance.}}

Solved Examples

Example 1: Easy (Flagstaff on a Building)

A flagstaff of height h is mounted on top of a 10 m high building. From a point P on the ground, the angle of elevation of the top of the building is 30°, and the angle of elevation of the top of the flagstaff is 45°. Find the height of the flagstaff. (Use √3 ≈ 1.732)

Given: Height of building AB = 10 m, ∠APB = 30°, ∠DPA = 45°.

To Find: Height of the flagstaff BD, and distance PA.

Solution:

1. Let the distance of the point P from the foot of the building be x meters, so PA = x. Let the height of the flagstaff BD be h meters. The total height of the flagstaff top from the ground is AD = AB + BD = 10 + h.

2. First, consider the smaller right-angled triangle, ΔPAB. The angle of elevation is 30°.

   tan 30° = AB / PA
   

   1/√3 = 10 / x
   

   x = 10√3 m
   

3. Now, consider the larger right-angled triangle, ΔPAD. The angle of elevation is 45°.

   tan 45° = AD / PA
   

   1 = (10 + h) / x
   

   x = 10 + h
   

4. We have two expressions for x. We can equate them to find h.

   10√3 = 10 + h
   

   h = 10√3 - 10
   

   h = 10(√3 - 1)
   

5. Substitute the value of √3 ≈ 1.732 to get the final answer.

   h = 10(1.732 - 1) = 10(0.732) = 7.32 m
   

Final Answer: The height of the flagstaff is 7.32 m.

Example 2: Medium (Observer Moving Towards an Object)

The angle of elevation of the top of a tower from a point on the ground is 30°. After walking 40 m towards the tower, the angle of elevation becomes 60°. Find the height of the tower.

Given: Initial angle of elevation = 30°, final angle of elevation = 60°, distance walked = 40 m.

To Find: The height of the tower.

{{VISUAL: diagram: A vertical tower AB of height 'h'. Two points on the ground, C and D, are in a straight line with B. The angle of elevation from C is 30° and from D is 60°. The distance CD is 40m. The distance DB is 'x'.}}

Solution:

1. Let AB be the tower of height h. Let C be the initial point and D be the final point of observation. Let DB = x. Then CB = CD + DB = 40 + x.

2. Consider the right-angled triangle ΔABD. The angle of elevation from D is 60°.

   tan 60° = AB / DB
   

   √3 = h / x
   

   h = x√3   ---(Equation 1)
   

3. Now, consider the right-angled triangle ΔABC. The angle of elevation from C is 30°.

   tan 30° = AB / CB
   

   1/√3 = h / (40 + x)
   

```
h√3 = 40 + x   ---(Equation 2)
```

4. Substitute the value of h from Equation 1 into Equation 2.

```
(x√3) × √3 = 40 + x
```

```
3x = 40 + x
```

```
2x = 40
```

```
x = 20 m
```

5. Now substitute the value of x back into Equation 1 to find the height h.

```
h = 20 × √3
```

```
h = 20√3 m
```

Final Answer: The height of the tower is 20√3 m.

Example 3: Hard (Angles of Elevation & Depression)

From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

Given: Building height = 7 m, angle of elevation to tower top = 60°, angle of depression to tower foot = 45°.

To Find: The height of the cable tower.

{{VISUAL: diagram: A vertical building AB (7m high) and a taller vertical tower CD. A horizontal line AE is drawn from A to CD. Angle of elevation ∠CAE is 60°. Angle of depression ∠EAD is 45°, which makes ∠ADB = 45° (alternate interior angles). BE=7m, BD is the distance 'x' between buildings.}}

Solution:

1. Let AB be the building of height 7 m and CD be the cable tower of height H. Draw a line AE parallel to the ground from point A, meeting CD at E. Thus, AE = BD (distance between them) and CE is the part of the tower above the building. Also, ED = AB = 7 m. The total height of the tower is H = CE + ED.

2. First, consider the lower triangle, ΔABD, which is right-angled at B. The angle of depression from A to D is 45°, so ∠EAD = 45°. Since AE is parallel to BD, ∠ADB = ∠EAD = 45° (alternate interior angles).

   tan 45° = AB / BD
   

   1 = 7 / BD
   

   BD = 7 m
   

   This means the distance between the building and the tower is 7 m. So, AE = 7 m.

3. Now, consider the upper triangle, ΔAEC, which is right-angled at E. The angle of elevation is 60°.

   tan 60° = CE / AE
   

   √3 = CE / 7
   

   CE = 7√3 m
   

4. The total height of the tower H is the sum of CE and ED.

   H = CE + ED
   

   H = 7√3 + 7
   

   H = 7(√3 + 1) m
   

Final Answer: The height of the tower is 7(√3 + 1) m.

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

1. The shadow of a vertical tower on level ground increases by 10 meters when the sun’s altitude changes from 45° to 30°. Find the height of the tower.

   💡 Answer: Let h be the height and x be the initial shadow length. tan 45° = h/x → h = x. tan 30° = h/(x+10) → h√3 = x+10. Substitute x=h: h√3 = h+10 → h(√3 - 1) = 10. h = 10 / (√3 - 1). Rationalizing gives h = 5(√3 + 1) meters.

2. Two poles of equal height are standing opposite each other on either side of a road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°. Find the height of the poles and the distances of the point from the poles.

   💡 Answer: Let h be the height, and the point be x meters from the first pole. It is (80-x) m from the second. tan 60° = h/x → h = x√3. tan 30° = h/(80-x) → h√3 = 80-x. Substitute h: (x√3)√3 = 80-x → 3x = 80-x → 4x = 80 → x = 20. The distances are 20 m and 60 m. The height is h = 20√3 meters.

3. The angle of elevation of a cloud from a point 60 m above a lake is 30° and the angle of depression of the reflection of the cloud in the lake is 60°. Find the height of the cloud from the surface of the lake.

   💡 Answer: This is a classic tricky problem! Let the lake surface be the ground line. The observer is at point A, 60m above the lake. Let the cloud be at point C, at height H above the lake. The reflection C' will be H meters below the lake surface. Let the horizontal distance be x. From the observer at A, the height to the cloud is (H-60). tan 30° = (H-60)/x. The depth to the reflection is (H+60). tan 60° = (H+60)/x. From the first eq, x = (H-60)√3. From the second, x = (H+60)/√3. Equate them: (H-60)√3 = (H+60)/√3 → 3(H-60) = H+60 → 3H - 180 = H + 60 → 2H = 240. The height of the cloud H is 120 meters.

Mini Cheatsheet

Heights and Distances — Part 3 (Complex Scenarios)

Heights and Distances — Part 3 (Complex Scenarios)

Welcome back! So far, we've tackled problems that can be represented by a single right-angled triangle. But the real world is rarely that simple. What if you're standing on a cliff and need to calculate the height of a lighthouse across the water? Or what if an airplane is flying towards you, and you want to calculate its speed based on how its angle of elevation changes?

These scenarios are more complex because they involve two or more right-angled triangles interacting within the same problem. The key to solving them is to break the complex picture down into simpler parts. You'll often find that the two triangles share a common side or have a side that can be related to the other. By setting up trigonometric equations for each triangle and solving them simultaneously, we can unlock the secrets of these more intricate real-world situations.

{{FORMULA: expr=tan θ = Opposite / Adjacent | symbols=θ:Angle of elevation or depression, Opposite:Side opposite to the angle, Adjacent:Side adjacent to the angle}}

Definitions & Formulas Recap

Before we dive into complex problems, let's refresh the core tools. All problems on this page will rely on these fundamental concepts.

The Two-Triangle Problem-Solving Logic

Solving problems with multiple triangles follows a consistent, logical process. The goal is to find a link between the triangles and use it to solve for the unknown.

1. Draw a Clear Diagram: This is the most crucial step. Read the problem carefully and translate the words into a geometric figure. Label all known lengths and angles. Mark the vertices (A, B, C, etc.) clearly.

2. Identify the Right-Angled Triangles: Look at your diagram and identify the two (or sometimes more) right-angled triangles that contain your knowns and unknowns.

3. Find the "Bridge": The triangles will be connected in some way. You must find this connection. It's usually:

   • A common side (e.g., the height of a tower is a side in both triangles).
   • A common vertex.
   • Sides that add up to a known length (e.g., parts of a road).

4. Formulate Equations: For each right-angled triangle, write a trigonometric equation using the ratio (tan, sin, or sin) that connects the known information to the unknown. You will now have a system of two equations.

5. Solve the System: Use algebraic methods like substitution or elimination to solve the two equations for the required unknown value. Often, you'll solve for the "bridge" side in one equation and substitute it into the other.

Solved Examples

Let's apply this logic to problems of increasing difficulty.

Example 1: The Flagstaff on a Building (Easy)

A 10 m tall building has a flagstaff on top. From a point P on the ground, the angle of elevation of the bottom of the flagstaff is 30°, and the angle of elevation of the top of the flagstaff is 45°. Find the height of the flagstaff. (Use √3 ≈ 1.732)

Given: Height of building (AB) = 10 m, Angle of elevation to bottom (∠APB) = 30°, Angle of elevation to top (∠APD) = 45°.

To Find: Height of the flagstaff (BD).

Solution:

1. Let the point on the ground be P, the base of the building be A, and the top of the building be B. Let the top of the flagstaff be D. The height of the flagstaff is BD. We have two right triangles: ΔPAB and ΔPAD. The common side is PA.

2. First, consider the smaller triangle, ΔPAB. We use the tan ratio to find the distance PA.

tan(30°) = AB / PA

3. Substitute the known values and solve for PA.

1/√3 = 10 / PA

PA = 10√3 m

4. Now, consider the larger triangle, ΔPAD. The total height AD is the sum of the building's height and the flagstaff's height (AD = AB + BD = 10 + h).

tan(45°) = AD / PA

5. Substitute the known values, including the value of PA we just found.

1 = (10 + BD) / (10√3)

6. Now, solve for the height of the flagstaff, BD.

10√3 = 10 + BD

BD = 10√3 - 10

BD = 10(√3 - 1)

7. Calculate the final value using √3 ≈ 1.732.

BD = 10(1.732 - 1) = 10(0.732) = 7.32 m

Final Answer: The height of the flagstaff is 7.32 m.

Example 2: Observing a Tower from a Building (Medium)

From the top of a 60 m high building, the angles of depression of the top and the bottom of a tower are observed to be 30° and 60° respectively. Find the height of the tower.

Given: Height of building (AB) = 60 m, Angle of depression to tower top (∠EAC) = 30°, Angle of depression to tower bottom (∠EAD) = 60°.

To Find: Height of the tower (CD).

{{VISUAL: diagram: A taller building on the left labeled AB, and a shorter tower on the right labeled CD. A horizontal line extends from A, the top of the building. Angles of depression are shown from this line down to C and D.}}

Solution:

1. Let AB be the building and CD be the tower. Let's draw a horizontal line AE from the top of the building. The distance between them is BD. Let's also draw a line CE parallel to BD. So, BD = CE and BE = CD. We have two right triangles: ΔAEC and ΔABD.

2. From the properties of parallel lines, the alternate interior angles are equal.

   • ∠ACB = ∠EAC = 30° (Angle of elevation from tower top to building top)
   • ∠ADB = ∠EAD = 60° (Angle of elevation from tower bottom to building top)

3. First, consider the larger triangle, ΔABD. We can find the distance between the buildings, BD.

tan(60°) = AB / BD

4. Substitute the values and solve for BD.

√3 = 60 / BD

BD = 60 / √3 = 20√3 m

5. Since BD = CE, we now know CE = 20√3 m.

6. Now, consider the other triangle, ΔAEC. The height AE is the height of the building minus the height of the part equal to the tower (AE = AB - BE). Since BE = CD (height of tower), AE = 60 - CD.

tan(30°) = AE / CE

7. Substitute the known values into this equation.

1/√3 = (60 - CD) / (20√3)

8. Solve for CD.

(20√3) / √3 = 60 - CD

20 = 60 - CD

CD = 60 - 20 = 40 m

Final Answer: The height of the tower is 40 m.

{{KEY: type=strategy | title=The Common Side is Key | text=In almost every two-triangle problem, the first step is to use one triangle to find the length of the side that is common to both triangles. This common side acts as a bridge, allowing you to transfer information from one triangle to the other to find your final answer.}}

Example 3: The Moving Car (Hard)

A man on top of a vertical tower observes a car moving at a uniform speed coming directly towards it. If it takes 12 minutes for the angle of depression to change from 30° to 45°, how soon after this will the car reach the tower?

Given: Time taken to change angle from 30° to 45° is 12 minutes.

To Find: Time taken for the car to travel from the 45° point to the base of the tower.

Solution:

1. Let AB be the tower of height h. Let C be the initial position of the car (angle of depression 30°) and D be its position after 12 minutes (angle of depression 45°). The car is moving towards the base of the tower, B. The angles of elevation from C and D to A are ∠ACB = 30° and ∠ADB = 45°.

2. Let the speed of the car be s. The distance covered in 12 minutes is CD. So, CD = s × 12. Let the time taken to travel from D to B be t minutes. So, DB = s × t.

3. Consider the right triangle ΔABD.

tan(45°) = AB / DB

1 = h / (s × t)  →  h = s × t

This is our first equation.

4. Now, consider the right triangle ΔABC. The base CB is CD + DB.

tan(30°) = AB / CB

1/√3 = h / (CD + DB)

1/√3 = h / (12s + st)

This is our second equation.

5. Now we substitute the value of h from the first equation (h = st) into the second equation.

1/√3 = (st) / (12s + st)

6. The speed s is common in the numerator and denominator, so it can be cancelled out.

1/√3 = (st) / (s(12 + t))

1/√3 = t / (12 + t)

7. Now, we solve this simple equation for t.

12 + t = √3 × t

12 = √3t - t

12 = t(√3 - 1)

t = 12 / (√3 - 1)

8. To simplify, we rationalize the denominator by multiplying the numerator and denominator by (√3 + 1).

t = [12(√3 + 1)] / [(√3 - 1)(√3 + 1)]

t = [12(√3 + 1)] / (3 - 1)

t = [12(√3 + 1)] / 2

t = 6(√3 + 1)

9. Using √3 ≈ 1.732, we can find the approximate time.

t = 6(1.732 + 1) = 6(2.732) ≈ 16.39 minutes

(For exam purposes, leaving the answer as 6(√3 + 1) is often acceptable unless specified otherwise.)

Final Answer: The car will reach the tower in 6(√3 + 1) minutes (approx. 16.39 minutes).

Example 4: The Shadow Problem (Tricky)

The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun's altitude (angle of elevation) is 30°, than when it is 60°. Find the height of the tower.

Given: Difference in shadow length between 30° and 60° elevation is 40 m.

To Find: Height of the tower.

Solution:

1. Let AB be the tower of height h. Let C be the tip of the shadow when the angle of elevation is 60° (∠ACB = 60°). Let D be the tip of the shadow when the angle is 30° (∠ADB = 30°). The length of the shadow at C is BC. The length of the shadow at D is BD. We are given that CD = 40 m.

2. Consider the right triangle ΔABC.

tan(60°) = AB / BC

√3 = h / BC  →  BC = h / √3

3. Now, consider the larger right triangle ΔABD. The base BD is BC + CD.

tan(30°) = AB / BD

1/√3 = h / (BC + 40)

4. We have two variables, h and BC. Let's substitute the expression for BC from step 2 into the equation from step 3.

1/√3 = h / ( (h/√3) + 40 )

5. Now, we solve this equation for h.

(h/√3) + 40 = h√3

6. Bring all terms with h to one side.

40 = h√3 - h/√3

7. Take h as a common factor.

40 = h (√3 - 1/√3)

8. Simplify the expression in the parenthesis.

40 = h ( (3 - 1) / √3 )

40 = h (2 / √3)

9. Isolate h to find the height of the tower.

h = (40 × √3) / 2

h = 20√3 m

Final Answer: The height of the tower is 20√3 m.

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

1. The angle of elevation of the top of a tower from two points at distances a and b from the base and in the same straight line with it are complementary. Prove that the height of the tower is √(ab).

   💡 Answer: Let the angles be θ and (90°-θ). Let h be the height. We get two equations: tan θ = h/a and tan(90°-θ) = h/b. Since tan(90°-θ) = cot θ = 1/tan θ, we have 1/tan θ = h/b. Substitute tan θ = h/a into this: 1/(h/a) = h/b → a/h = h/b → h² = ab → h = √(ab).

2. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

   💡 Answer: The distance to the tower is found from the depression angle: tan(45°) = 7 / distance → distance = 7 m. The height of the tower above the building is found from the elevation angle: tan(60°) = height_above / 7 → height_above = 7√3 m. Total tower height = 7 + 7√3 = 7(1 + √3) m.

3. An aeroplane when flying at a height of 4000 m from the ground passes vertically above another aeroplane at an instant when the angles of the elevation of the two planes from the same point on the ground are 60° and 45° respectively. Find the vertical distance between the aeroplanes at that instant.

   💡 Answer: Let the point on the ground be P. Let the lower plane be at B and the higher one at A. Let the point on the ground directly below them be C. We have two triangles, ΔPCB and ΔPCA. For the lower plane: tan(45°) = BC / PC → 1 = BC / PC → PC = BC. For the higher plane: tan(60°) = AC / PC → √3 = 4000 / PC → PC = 4000/√3. Therefore, BC = 4000/√3. The vertical distance is AB = AC - BC = 4000 - 4000/√3 = 4000(1 - 1/√3) m.

Mini Cheatsheet

Summary & Quick Revision

{{FORMULA: expr=SOH CAH TOA | symbols=sin θ:Opposite/Hypotenuse, cos θ:Adjacent/Hypotenuse, tan θ:Opposite/Adjacent}}

Chapter Summary & Quick Revision

Welcome to the final review of our journey into the practical world of trigonometry! This chapter was all about bridging the gap between abstract trigonometric ratios and the real world. We've learned how to measure the unmeasurable—calculating the height of a towering skyscraper, the width of a flowing river, or the distance to a ship at sea, all from a single vantage point. By mastering the concepts of line of sight, angle of elevation, and angle of depression, we've equipped ourselves with the tools of ancient astronomers and modern engineers. This summary will crystallize those concepts, ensuring you can tackle any height and distance problem with confidence.

Core Definitions

At the heart of this chapter are a few key terms and the fundamental trigonometric ratios. Understanding these is non-negotiable for solving problems.

The Problem-Solving Framework

Every problem in this chapter, no matter how complex it seems, can be broken down into a series of logical steps. This framework turns word problems into simple right-angled triangles that you already know how to solve.

1. Read and Visualize: Read the problem carefully, twice if needed. Draw a rough sketch representing the situation. This is the most crucial step. Label all points (e.g., A, B, C) and mark the right angle.

2. Identify and Label: Mark the given information on your diagram—lengths, distances, and angles. Clearly identify what you need to find. Your diagram should visually represent the knowns and the unknown.

{{VISUAL: diagram: A diagram showing a horizontal line from an observer's eye. The line of sight points upwards to the top of a tree, forming the angle of elevation. Another line of sight points downwards to a ball on the ground, forming the angle of depression.}}

3. Select the Right Tool: Look at the angle you have and the sides you're working with (the known side and the unknown side). Decide which trigonometric ratio connects them:

   • Use sin θ if you have Opposite and Hypotenuse.
   • Use cos θ if you have Adjacent and Hypotenuse.
   • Use tan θ if you have Opposite and Adjacent. (Hint: This is the most commonly used ratio in this chapter.)

4. Formulate and Solve: Write the equation using the chosen ratio. Substitute the known values and solve for the unknown variable. Remember to rationalize the denominator if it contains a square root.

{{KEY: type=concept | title=The Diagram is Everything | text=In 'Applications of Trigonometry', a well-drawn and correctly labeled diagram is 50% of the solution. It transforms a complex word problem into a simple geometric puzzle. Never skip this step!}}

Solved Examples

Let's apply the framework to problems of increasing difficulty.

Example 1: The Leaning Ladder (Easy)

Given: A ladder 10 m long reaches a window 5 m high from the ground.

To Find: The angle the ladder makes with the ground.

Solution:

1. Let AC be the ladder, AB be the height of the window, and BC be the distance from the wall. We have a right-angled triangle ABC, with the right angle at B.

2. Here, the Hypotenuse (AC) is 10 m and the side Opposite to the angle θ (AB) is 5 m.

3. We need a ratio that connects Opposite and Hypotenuse. That's the sin ratio.

sin θ = Opposite / Hypotenuse = AB / AC

4. Substitute the given values into the equation.

sin θ = 5 / 10 = 1/2

5. We know that sin 30° = 1/2. Therefore, the angle is 30°.

θ = 30°

Final Answer: The angle the ladder makes with the ground is 30°.

Example 2: The Approaching Boat (Medium)

Given: From the top of a 100 m high lighthouse, the angle of depression of a boat is 30°. After some time, the angle of depression becomes 60°.

To Find: The distance travelled by the boat during this period. (Use √3 = 1.73)

Solution:

1. Let AB be the lighthouse. Let C and D be the two positions of the boat. The angle of depression from A to C is 30° and to D is 60°. This means ∠PAC = 30° and ∠PAD = 60°. By alternate interior angles, ∠ACB = 30° and ∠ADB = 60°.

2. We have two right-angled triangles: ΔABD and ΔABC. The height AB = 100 m. We need to find the distance CD.

3. First, consider the triangle ΔABD. We use the tan ratio to find BD.

tan 60° = AB / BD

√3 = 100 / BD   →   BD = 100 / √3

4. Next, consider the triangle ΔABC. We use the tan ratio to find the total distance BC.

tan 30° = AB / BC

1/√3 = 100 / BC   →   BC = 100√3

5. The distance travelled by the boat is CD, which is BC - BD.

CD = 100√3 - (100 / √3)

6. Simplify the expression.

CD = (300 - 100) / √3 = 200 / √3

7. Rationalize the denominator and substitute the value of √3.

CD = (200√3) / 3 = (200 × 1.73) / 3 = 346 / 3 ≈ 115.33 m

Final Answer: The distance travelled by the boat is approximately 115.33 m.

Example 3: The Broken Tree (Hard)

Given: A tree breaks due to a storm. The broken part bends so that the top of the tree touches the ground, making an angle of 30°. The distance from the foot of the tree to the point where the top touches the ground is 8 m.

To Find: The original height of the tree.

Solution:

1. Let the tree be AB. It breaks at point C. The broken part AC' bends to touch the ground at D. So, AC' = CD. The original height of the tree is BC + CD. We have a right-angled triangle ΔCBD, with ∠CDB = 30° and BD = 8 m.

2. First, we find the height of the unbroken part, BC, using the tan ratio.

tan 30° = BC / BD

1/√3 = BC / 8   →   BC = 8 / √3 m

3. Next, we find the length of the broken part, CD, using the cos ratio.

cos 30° = BD / CD

√3 / 2 = 8 / CD   →   CD = 16 / √3 m

4. The total height of the tree is the sum of the unbroken part (BC) and the broken part (CD).

Total Height = BC + CD = (8 / √3) + (16 / √3)

Total Height = 24 / √3

5. Rationalize the denominator to get the final answer.

Total Height = (24 × √3) / (√3 × √3) = 24√3 / 3 = 8√3 m

Final Answer: The original height of the tree was 8√3 m.

Example 4: The Car and the Tower (Tricky)

Given: A car is moving towards a tower at a uniform speed. From the top of the tower, the angle of depression to the car is 30°. Six seconds later, the angle of depression is 60°.

To Find: The time taken by the car to reach the foot of the tower from the second point.

Solution:

1. Let AB be the tower of height h. Let C be the initial position of the car and D be its position after 6 seconds. The angles of elevation from C and D to A are 30° and 60° respectively.

2. In right-angled ΔABD, we use tan to relate h and BD.

tan 60° = AB / BD = h / BD

√3 = h / BD   →   h = BD√3   (Equation 1)

3. In right-angled ΔABC, we use tan to relate h and BC. Note that BC = BD + DC.

tan 30° = AB / BC = h / (BD + DC)

1/√3 = h / (BD + DC)   →   BD + DC = h√3   (Equation 2)

4. Substitute the value of h from Equation 1 into Equation 2.

BD + DC = (BD√3) × √3

BD + DC = 3BD

DC = 2BD   or   BD = DC / 2

5. This result is key: the distance BD is half the distance DC. We know the car took 6 seconds to travel distance DC.

6. Since the speed is uniform, the time taken is proportional to the distance.

   • Time to travel DC = 6 seconds.
   • Time to travel BD = Time to travel DC / 2.

Time for BD = 6 seconds / 2 = 3 seconds

Final Answer: The car will take 3 seconds to reach the foot of the tower from the second point.

Tips & Tricks

Common Mistakes to Avoid

Brain-Teaser Questions

1. The Shadow Problem: The shadow of a tower is x meters long when the sun's angle of elevation is 60°. When the sun's angle of elevation drops to 30°, the shadow becomes 40 meters longer. What is the height of the tower?

   💡 Answer: Let h be the height. In the first case, tan 60° = h / x → √3 = h / x → h = x√3. In the second case, tan 30° = h / (x + 40) → 1/√3 = h / (x + 40) → x + 40 = h√3. Substitute h from the first equation into the second: x + 40 = (x√3)√3 → x + 40 = 3x → 2x = 40 → x = 20 m. Now find h: h = x√3 = 20√3 m. The height of the tower is 20√3 meters.

2. The Two Observers: Two people are standing on opposite sides of a 50 m high tower. They observe the angle of elevation of the top of the tower to be 30° and 60° respectively. What is the distance between the two people?

   💡 Answer: Let the tower be AB = 50 m. Let the observers be at C and D on opposite sides. For observer at C: tan 30° = 50 / BC → 1/√3 = 50 / BC → BC = 50√3 m. For observer at D: tan 60° = 50 / BD → √3 = 50 / BD → BD = 50/√3 m. The total distance CD = BC + BD = 50√3 + 50/√3 = (150 + 50)/√3 = 200/√3 m. The distance between them is (200√3)/3 meters.

3. The Flying Bird: A bird is flying at a constant height of 1200 m. From a point on the ground, its angle of elevation is 60°. After 10 seconds, from the same point, its angle of elevation is 30°. Find the speed of the bird in m/s.

   💡 Answer: Let the initial position of the bird be A and final be B. The observer is at O. Height h = 1200 m. Initial horizontal distance OC: tan 60° = 1200 / OC → √3 = 1200 / OC → OC = 1200/√3 = 400√3 m. Final horizontal distance OD: tan 30° = 1200 / OD → 1/√3 = 1200 / OD → OD = 1200√3 m. Distance flown by bird CD = OD - OC = 1200√3 - 400√3 = 800√3 m. Time taken = 10 seconds. Speed = Distance / Time = (800√3) / 10 = 80√3 m/s. The speed is 80√3 m/s.

Mini Cheatsheet