Hello, students! Welcome to today's mathematics lesson. I'm so happy to be here with you to explore an exciting chapter — Chapter 9: Some Applications of Trigonometry. Now, in your previous chapter, you have already learned about trigonometric ratios like sine, cosine, and tangent. Today, we will see how these ratios are not just abstract mathematical concepts but are actually used to solve real-life problems. Are you ready to discover how trigonometry helps us measure heights and distances that we cannot measure directly? Let's begin!
Imagine you are standing on the ground and looking up at the top of a tall minar or a tree. Have you ever wondered how tall that minar actually is? You cannot simply take a ruler and measure it, can you? This is where trigonometry comes to our rescue. In this chapter, we will learn how to calculate heights of buildings, towers, and flagpoles, as well as distances between objects, just by knowing some angles and some simple measurements on the ground.
Now, before we start solving problems, let me introduce you to some very important terms that you will be using throughout this chapter.
The first term is the line of sight. When you look at an object, the line that goes from your eye to the point you are looking at is called the line of sight. Simple, isn't it? Now, when you raise your head to look at something that is above your eye level, the angle formed between this line of sight and the horizontal line is called the angle of elevation. Think about it — when you look up at the top of a tree or a building, you are raising your head, and the angle your eyes make with the horizontal is the angle of elevation. This is a very important concept, so remember: angle of elevation happens when the object is above your eye level.
Now, what happens when the object is below your eye level? Imagine you are sitting on a balcony and looking down at a flower pot on the ground. In this case, you are lowering your head to look at the object. The angle formed between your line of sight and the horizontal, but this time below the horizontal, is called the angle of depression. So, angle of depression is used when you are looking downwards at something.
Let me make this clearer with a simple example. Suppose you are standing on the roof of a building and looking at a car parked on the road below. The angle your eyes make with the horizontal while looking down at the car is the angle of depression. On the other hand, if you are standing on the ground and looking at the top of a flagpole, the angle your eyes make with the horizontal while looking up is the angle of elevation. These two terms will be the foundation of everything we do in this chapter, so make sure you understand them well.
Now that we know these terms, let us understand how to solve problems using them. Consider a situation where there is a tower or a minar, and you want to find its height without actually measuring it. What information would you need? You would need three things: first, the distance from where you are standing to the foot of the tower; second, the angle of elevation from your position to the top of the tower; and third, your own height, because your eyes are not at ground level — they are at some height above the ground. Once you have these three pieces of information, you can use trigonometric ratios to find the height of the tower.
Let me explain this step by step. Suppose you are standing at a point on the ground, and the distance from you to the foot of the tower is known. You measure the angle of elevation to the top of the tower. Now, if you draw a diagram, you will have a right-angled triangle where the tower forms one side, the ground distance forms the base, and the line of sight forms the hypotenuse. By using the tangent ratio, which is opposite over adjacent, you can find the vertical height of the tower above your eye level. Then, by adding your own height to this value, you get the total height of the tower. This is the basic approach we will use for all problems in this chapter.
Now, let me show you how this works with actual examples. We will go through each example carefully so that you understand the method completely.
Let's start with Example 1. A tower stands vertically on the ground. From a point on the ground, which is 15 meters away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60 degrees. We need to find the height of the tower.
First, let us draw a diagram to visualize the problem. We have a tower AB, which is vertical, so it makes a right angle with the ground at point B. The point from where we are observing is point C, which is 15 meters away from the foot of the tower, so BC equals 15 meters. The angle of elevation from point C to the top of the tower is 60 degrees, which is angle ACB. We need to find the height of the tower, which is AB.
Notice that triangle ACB is a right-angled triangle, right-angled at B. We know the angle at C is 60 degrees, and we know the base BC is 15 meters. We need to find the vertical side AB. Which trigonometric ratio should we use? Since we know the angle and the base, and we need to find the height, we can use the tangent ratio. Tan of an angle is opposite over adjacent. Here, for angle C, the opposite side is AB, and the adjacent side is BC. So, tan 60 degrees equals AB divided by BC. We know tan 60 degrees is the square root of 3. So, square root of 3 equals AB divided by 15. Solving this, we get AB equals 15 times square root of 3. Therefore, the height of the tower is 15 square root of 3 meters. That's approximately 25.98 meters, but we usually leave it in the exact form as 15√3 meters.
Great! Now let's move to Example 2. This is a very practical problem. An electrician has to repair an electric fault on a pole of height 5 meters. She needs to reach a point 1.3 meters below the top of the pole to undertake the repair work. What should be the length of the ladder that she should use, which when inclined at an angle of 60 degrees to the horizontal, would enable her to reach the required position? Also, how far from the foot of the pole should she place the foot of the ladder? We are told to take √3 as 1.73.
Let's understand the problem. We have a pole AD, which is 5 meters tall. The electrician needs to reach point B, which is 1.3 meters below the top, so the distance from the ground to point B is 5 minus 1.3, which equals 3.7 meters. She will use a ladder BC, which makes an angle of 60 degrees with the horizontal ground. We need to find the length of the ladder, which is BC, and also how far from the pole she should place the ladder, which is DC.
In the right triangle BDC, we know that angle C is 60 degrees, and we know the vertical side BD, which is 3.7 meters. We need to find the hypotenuse BC, which is the length of the ladder. Which ratio should we use? Since we know the vertical side and we need the hypotenuse, we should use the sine ratio. Sin of an angle is opposite over hypotenuse. So, sin 60 degrees equals BD divided by BC. We know sin 60 degrees is √3/2. So, √3/2 equals 3.7 divided by BC. Solving this, BC equals 3.7 times 2 divided by √3, which is 7.4 divided by √3. Using √3 as 1.73, this is approximately 7.4 divided by 1.73, which is about 4.28 meters. So, the ladder should be approximately 4.28 meters long.
Now, to find how far from the pole she should place the ladder, we need to find DC. In the same right triangle, we can use the cotangent ratio. Cot of an angle is adjacent over opposite. So, cot 60 degrees equals DC divided by BD. We know cot 60 degrees is 1/√3. So, 1/√3 equals DC divided by 3.7. Therefore, DC equals 3.7 divided by √3, which is approximately 3.7 divided by 1.73, which is about 2.14 meters. So, she should place the foot of the ladder at a distance of approximately 2.14 meters from the pole.
Now, let's look at Example 3. An observer who is 1.5 meters tall is 28.5 meters away from a chimney. The angle of elevation of the top of the chimney from her eyes is 45 degrees. We need to find the height of the chimney.
In this problem, we have a chimney AB, and the observer is represented by CD, with her eyes at point D. The distance from the observer to the chimney is 28.5 meters, which is DE or CB. The observer's height is 1.5 meters, which is CE or DB. The angle of elevation from the observer's eyes to the top of the chimney is 45 degrees, which is angle ADE.
We need to find the total height of the chimney, which is AB. This height is equal to AE plus EB. Now, EB is the height of the observer, which is 1.5 meters. So, we need to find AE. In the right triangle ADE, right-angled at E, we know the angle at D is 45 degrees, and we know DE is 28.5 meters. We need to find AE. Using the tangent ratio, tan 45 degrees equals AE divided by DE. We know tan 45 degrees is 1. So, 1 equals AE divided by 28.5, which means AE equals 28.5 meters. Therefore, the total height of the chimney AB equals AE plus EB, which is 28.5 plus 1.5, equals 30 meters. So, the chimney is 30 meters tall.
Now, let's move to Example 4. This is a slightly more complex problem because it involves two right triangles. From a point P on the ground, the angle of elevation of the top of a 10-meter tall building is 30 degrees. A flag is hoisted at the top of the building, and the angle of elevation of the top of the flagstaff from point P is 45 degrees. We need to find the length of the flagstaff and the distance of the building from point P. We are told to take √3 as 1.732.
Let's understand this. We have a building AB, which is 10 meters tall. At the top of the building, there is a flagstaff BD. Point P is the point on the ground from where we are observing. We know that the angle of elevation from P to the top of the building is 30 degrees, and the angle of elevation from P to the top of the flagstaff is 45 degrees. We need to find the length of the flagstaff, which is BD, and the distance from P to the building, which is PA.
First, let's find the distance PA. In the right triangle PAB, we know AB is 10 meters, and angle APB is 30 degrees. Using the tangent ratio, tan 30 degrees equals AB divided by AP. We know tan 30 degrees is 1/√3. So, 1/√3 equals 10 divided by AP. Solving this, AP equals 10 times √3, which is 10√3 meters. Using √3 as 1.732, this is approximately 17.32 meters. So, the building is about 17.32 meters away from point P.
Now, we need to find the length of the flagstaff BD. Let BD be x meters. Then, the total height from the ground to the top of the flagstaff is AD, which is AB plus BD, that is 10 plus x meters. Now, consider the right triangle PAD. We know the angle of elevation to the top of the flagstaff is 45 degrees, so tan 45 degrees equals AD divided by AP. We know tan 45 degrees is 1. So, 1 equals (10 + x) divided by AP, which is (10 + x) divided by 10√3. Solving this, we get 10 + x equals 10√3. Therefore, x equals 10√3 minus 10, which is 10 times (√3 minus 1). Using √3 as 1.732, this is 10 times (1.732 minus 1), which is 10 times 0.732, which is 7.32 meters. So, the length of the flagstaff is approximately 7.32 meters.
Now, let's look at Example 5. This is an interesting problem about shadows. The shadow of a tower standing on level ground is found to be 40 meters longer when the Sun's altitude is 30 degrees than when it is 60 degrees. We need to find the height of the tower.
Let's understand this. When the Sun is at a certain altitude, the light rays hit the tower and cast a shadow on the ground. The angle of elevation of the Sun is the same as the angle of elevation of the top of the tower from the tip of its shadow. So, when the Sun's altitude is 60 degrees, the shadow of the tower is shorter. When the Sun's altitude is 30 degrees, the shadow is longer. We are told that the longer shadow is 40 meters more than the shorter shadow.
Let AB be the tower, with height h meters. Let BC be the length of the shadow when the Sun's altitude is 60 degrees. Let DB be the length of the shadow when the Sun's altitude is 30 degrees. According to the problem, DB is 40 meters longer than BC. So, DB equals x plus 40, where BC is x meters.
Now, consider the right triangle ABC, where angle ACB is the angle of elevation of the Sun, which is 60 degrees. Using the tangent ratio, tan 60 degrees equals AB divided by BC, which is h divided by x. We know tan 60 degrees is √3. So, √3 equals h divided by x, which means h equals x√3. This is equation (1).
Now, consider the right triangle ABD, where angle ADB is the angle of elevation of the Sun, which is 30 degrees. Using the tangent ratio, tan 30 degrees equals AB divided by BD, which is h divided by (x + 40). We know tan 30 degrees is 1/√3. So, 1/√3 equals h divided by (x + 40), which means h equals (x + 40) divided by √3. This is equation (2).
Now, we have two expressions for h. From equation (1), h equals x√3. From equation (2), h equals (x + 40)/√3. Since both equal h, we can equate them: x√3 equals (x + 40)/√3. Multiplying both sides by √3, we get 3x equals x + 40. Solving this, 3x minus x equals 40, so 2x equals 40, which means x equals 20. So, the shorter shadow is 20 meters long. Now, from equation (1), h equals x√3, which is 20√3 meters. So, the height of the tower is 20√3 meters, which is approximately 34.64 meters.
Now, let's look at Example 6. This problem involves angles of depression from one building to another. The angles of depression of the top and the bottom of an 8-meter tall building from the top of a multi-storeyed building are 30 degrees and 45 degrees, respectively. We need to find the height of the multi-storeyed building and the distance between the two buildings.
Let's understand the problem. We have a multi-storeyed building PC, and a shorter building AB, which is 8 meters tall. From the top of the multi-storeyed building, which is point P, we look down at the top and bottom of the shorter building. The angle of depression to the top of the shorter building is 30 degrees, and to the bottom is 45 degrees. We need to find the height of the multi-storeyed building, which is PC, and the distance between the two buildings, which is AC.
Let's draw the diagram carefully. We have point P at the top of the multi-storeyed building. The shorter building AB is 8 meters tall, so AB equals 8 meters. The distance between the buildings is AC. From point P, we draw lines to the top and bottom of the shorter building. Now, here is an important geometric fact: since the lines PQ and BD are parallel (they are both horizontal), the angle of depression from P to the top of the shorter building equals the angle of elevation from the bottom of the shorter building to P. These are alternate interior angles. So, angle PBD equals 30 degrees. Similarly, angle PAC equals 45 degrees.
Now, in the right triangle PBD, right-angled at D, we have angle PBD equals 30 degrees. We know that PD is the vertical height from the bottom of the shorter building to the top of the multi-storeyed building. Using the tangent ratio, tan 30 degrees equals PD divided by BD. We know tan 30 degrees is 1/√3. So, 1/√3 equals PD divided by BD, which means BD equals PD times √3. This is equation (1).
In the right triangle PAC, right-angled at C, we have angle PAC equals 45 degrees. Using the tangent ratio, tan 45 degrees equals PC divided by AC. We know tan 45 degrees is 1. So, 1 equals PC divided by AC, which means PC equals AC. This is equation (2).
Now, we also know that PC equals PD plus DC, where DC is the height of the shorter building, which is 8 meters. So, PC equals PD plus 8. And from equation (2), AC equals PC. Also, from the geometry, AC equals BD, because both are the distances between the buildings. So, we have BD equals AC, which equals PC.
Now, from equation (1), BD equals PD times √3. And we know BD equals PC. So, PC equals PD times √3. But PC also equals PD plus 8. So, PD plus 8 equals PD times √3. Solving this, PD times √3 minus PD equals 8, which is PD times (√3 minus 1) equals 8. So, PD equals 8 divided by (√3 minus 1). To simplify this, we multiply the numerator and denominator by (√3 + 1), which gives us PD equals 8 times (√3 + 1) divided by (3 minus 1), which is 8 times (√3 + 1) divided by 2, which is 4 times (√3 + 1). So, PD equals 4(√3 + 1) meters.
Now, the height of the multi-storeyed building PC equals PD plus 8, which is 4(√3 + 1) plus 8. Simplifying, this is 4√3 + 4 + 8, which is 4√3 + 12, which is 4(3 + √3) meters. Using √3 as 1.73, this is approximately 4 times (3 + 1.73), which is 4 times 4.73, which is 18.92 meters. And the distance between the two buildings, which is AC, equals PC, which is also 4(3 + √3) meters, approximately 18.92 meters.
Now, let's look at Example 7. This is a problem about a river. From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30 degrees and 45 degrees, respectively. If the bridge is at a height of 3 meters from the banks, we need to find the width of the river.
Let's understand this. We have a bridge, and from a point P on the bridge, we look down at the banks on either side of the river. The bridge is 3 meters above the banks, so PD equals 3 meters. The bank on one side is at point A, and the bank on the other side is at point B. The width of the river is AB, which is the distance between the two banks. We need to find AB.
From point P, the angle of depression to bank A is 30 degrees, and to bank B is 45 degrees. Now, remember that the angle of depression equals the angle of elevation from the bank to the point on the bridge, because the lines are parallel. So, angle APD equals 30 degrees, and angle BPD equals 45 degrees.
In the right triangle APD, right-angled at D, we have angle A equals 30 degrees. Using the tangent ratio, tan 30 degrees equals PD divided by AD. We know tan 30 degrees is 1/√3. So, 1/√3 equals 3 divided by AD, which means AD equals 3√3 meters.
In the right triangle BPD, right-angled at D, we have angle B equals 45 degrees. Using the tangent ratio, tan 45 degrees equals PD divided by BD. We know tan 45 degrees is 1. So, 1 equals 3 divided by BD, which means BD equals 3 meters.
Now, the width of the river AB equals AD plus DB, which is 3√3 plus 3 meters. We can factor out 3, so this is 3 times (√3 + 1) meters. Using √3 as 1.73, this is approximately 3 times (1.73 + 1), which is 3 times 2.73, which is 8.19 meters. So, the width of the river is approximately 8.19 meters.
Now, students, we have covered all the worked examples from the chapter. Let me summarize what we have learned in this chapter.
In this chapter, we have studied how trigonometry is used to solve real-life problems involving heights and distances. We learned three important concepts: the line of sight, which is the line from the observer's eye to the object being viewed; the angle of elevation, which is the angle formed when the object is above the observer's eye level; and the angle of depression, which is the angle formed when the object is below the observer's eye level.
We learned that by knowing the angle of elevation or depression and one side of a right triangle, we can use trigonometric ratios like sine, cosine, and tangent to find the other sides. We solved various types of problems, including finding the height of a tower, the length of a ladder, the height of a chimney, the length of a flagstaff, the height using shadows, and the width of a river. In all these problems, we carefully identified the right triangles, determined which trigonometric ratio to use, and solved the equations step by step.
The key points to remember are: always draw a clear diagram to visualize the problem; identify the right-angled triangle and the known quantities; choose the appropriate trigonometric ratio based on what sides you know and what you need to find; solve the equation carefully; and don't forget to add the observer's height when calculating the total height of an object.
Trigonometry is a powerful tool that helps us measure things that we cannot measure directly. Whether it's finding the height of a building, the length of a ladder needed to reach a certain point, or the width of a river, trigonometric ratios provide us with the solution. I hope you now have a clear understanding of how to apply trigonometry to solve such problems. Keep practicing, and you will become very comfortable with these concepts. Thank you for listening, and I'll see you in the next lesson!