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Today, we will explore how we can use trigonometric ratios to solve problems involving heights and distances. Can anyone tell me what trigonometric ratios we might use?
Sine, cosine, and tangent?
Exactly! Sine is opposite over hypotenuse, cosine is adjacent over hypotenuse, and tangent is opposite over adjacent. Let's look at our first problem about a circus artist climbing a pole.
How would we start solving the height of the pole?
Great question! Since we have the angle with respect to the ground, we can use the tangent function. In this scenario, if the angle is 30 degrees and the length of the rope is 20 meters, our height can be found using: height = length Γ sin(angle).
So, is the height sin(30Β°) times 20?
Correct! Remember that sin(30Β°) equals 0.5. Thus, the height is 20 Γ 0.5, which is 10 meters.
So the pole is 10 meters tall. Key points to remember are the trigonometric ratios and how they apply to real-life situations. Letβs move on!
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Now letβs discuss the scenario where a tree has broken and makes a 30Β° angle with the ground. What do we need to find?
We need to find the height of the tree, right?
Exactly! We know the distance from the base to the top of the tree touching the ground is 8 meters. To find the height, we again use the sine function. Can anyone suggest how to set up the equation?
So, height would be 8 Γ sin(30Β°)?
Correct! And since sin(30Β°) is 0.5, the height would be 8 Γ 0.5, resulting in 4 meters.
This exercise exemplifies how we can visualize trees and poles in our surroundings to apply mathematical concepts effectively.
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Letβs look at another problem where we compute the height of a tower based on angles of elevation measured from two different points. Who can summarize how to calculate this?
We could use the tangent for each angle to find two heights and then subtract or add them depending on the layout, right?
Yes! For example, if the angles of elevation are 30Β° and 60Β° from two different points, we can calculate the height based on the distance to the tower as 30 meters. We'll use the tangent function to derive the two different heights.
This sounds more complex! Could you explain how to manage both angles?
Certainly! First, weβll calculate the height from the base of the tower at a distance of 30 meters using both angles. For 30Β°, the height is 30 Γ tan(30Β°), and for 60Β°, it will be 30 Γ tan(60Β°). This lets us find the total height of the tower above the building.
Always remember, setting the scenarios correctly helps break down larger problems into manageable pieces. Letβs practice a few more examples.
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So far, weβve solved various problems. How should we approach a new scenario where the angle of elevation and distance to the top of a pole are given?
We would determine which ratio to use based on what weβve been given.
Absolutely correct! Letβs take a moment to analyze the question. For instance, if we're given a kite flying at a height of 60 meters with a string making a 60Β° angle, how will we calculate the length of the string?
We would use the cosine or sine based on the angle and opposite height!
Yes! It might be sine since we need the opposite side. So the string length can be determined as height divided by sine.
This emphasizes the significance of understanding relationships and thinking critically before computation.
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The exercises in this section explore real-world applications of trigonometry, specifically involving angles of elevation and depression. They require students to calculate heights of objects like poles, towers, and trees using given angles and distances.
This section focuses on practical applications of trigonometry, particularly through exercises that require calculations involving angles of elevation and depression. Each problem presents a scenario where students must determine unknown heights or distances using given lengths and angles. Important concepts include:
Through these exercises, learners gain insights into how trigonometry applies to everyday situations and the process of modeling real-world problems mathematically.
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In this scenario, we can visualize the situation as a right triangle where:
- The length of the rope is the hypotenuse (20 m).
- The height of the pole represents the opposite side to the angle of elevation (30Β°).
Using the sine function, which relates the opposite side and the hypotenuse, we apply the following formula:
\[ \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \]
where \( \theta = 30Β° \), the opposite side is the height of the pole (h), and the hypotenuse is 20 m. Therefore, we can rearrange the equation to find h:
\[ h = 20 \times \sin(30Β°) \]
Knowing that \( \sin(30Β°) = 0.5 \):
\[ h = 20 \times 0.5 = 10 \text{ m} \]
Thus, the height of the pole is 10 meters.
Imagine a circus artist climbing a 20m rope that is fixed like a straight line from the top of a tall pole to the ground. The angle of 30Β° formed with the ground is similar to looking up at a tall building. Just as you would use a measuring tape to find out how tall the building is by looking at how much you raise your arm upwards, we use mathematics to find out how tall the pole is based on the angle and the length of the rope.
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We can again visualize this as a right triangle. The broken tree acts like the hypotenuse, which creates an angle of 30Β° with the ground when it bent to touch the ground. Here, the distance from the base of the tree to where the top touches the ground is 8 m (the adjacent side). To find the height of the tree (the opposite side), we use the cosine function:
\[ \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} \]
With \( \theta = 30Β° \) and the adjacent side of 8 m, we rearrange the formula:
\[ \text{hypotenuse} = \frac{\text{adjacent}}{\cos(30Β°)} \]
Knowing that \( \cos(30Β°) = \frac{\sqrt{3}}{2} \), we find:
\[ \text{hypotenuse} = \frac{8}{\frac{\sqrt{3}}{2}} = \frac{16}{\sqrt{3}} \approx 9.24 m \]
The height of the tree is this length plus the portion above the break, which we calculate considering the right triangle formed by the total height and the broken part.
Think about a tree that gets bent during a storm, rather like a straw bending when pushed too far. If you look at it from the side while the top of the straw (the tree) touches the ground at an angle, you can use trigonometry to find how high it originally was, just like finding how tall something was before it fell over. It's like being a detective! You're using clues (like the angles and distances) to uncover the height of the tree.
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For the slide for younger children, we again have a right triangle. The height is 1.5 m (the opposite side) and the angle with the ground is 30Β°. We can find the hypotenuse (length of the slide) using sine:
\[ \sin(30Β°) = \frac{1.5}{\text{hypotenuse}} \]
Rearranging gives us:
\[ \text{hypotenuse} = \frac{1.5}{\sin(30Β°)} = \frac{1.5}{0.5} = 3 m \]
For the older childrenβs slide at a height of 3 m and an angle of 60Β°:
\[ \sin(60Β°) = \frac{3}{\text{hypotenuse}} \]
Rearranging gives us:
\[ \text{hypotenuse} = \frac{3}{\sin(60Β°)} = \frac{3}{\frac{\sqrt{3}}{2}} = \frac{6}{\sqrt{3}} \approx 3.46 m \]
Thus, the lengths of the slides for the younger and older children are 3 m and approximately 3.46 m, respectively.
Picture setting up slides at a playground. For younger kids, you want a gentle slope, not too tall and easy to climb, while for older kids, you might want a steep slide that provides a little thrill! By thinking about the height you want and the angle you have in mind, you can calculate just how long the slide needs to be. This is like planning the perfect roller coaster ride for different ages!
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For this problem, we identify a right triangle where:
- The distance from the point on the ground to the tower is the base (30 m).
- The height of the tower is the opposite side to the angle of elevation (30Β°). Understanding this, we can again use the tangent function, which is the ratio of the opposite side (height) to the adjacent side (30 m):
\[ \tan(30Β°) = \frac{h}{30} \]
Where \( \tan(30Β°) = \frac{1}{\sqrt{3}} \), we rearrange:
\[ h = 30 \cdot \tan(30Β°) = 30 \cdot \frac{1}{\sqrt{3}} \approx 17.32 ext{ m} \]
Thus, the height of the tower is approximately 17.32 m.
Imagine you stand back from a tall building and look up at the top to see how high it is. The closer you get, the better an angle you can create to see the top. By knowing how far away you are and the angle at which you look, math helps you discover the height of that building, just like calculating how tall the tower is from a certain distance!
Learn essential terms and foundational ideas that form the basis of the topic.
Key Concepts
Trigonometric Ratios: Used to find missing side lengths of triangles based on angles.
Angle of Elevation: Important in determining height when looking up.
Angle of Depression: Vital for calculating height when looking down.
See how the concepts apply in real-world scenarios to understand their practical implications.
A circus artist climbs a rope at a 30Β° angle, allowing us to find the height of the pole using trigonometric functions.
A broken tree creates an angle with the ground; using distance to the tree base helps find its height.
Use mnemonics, acronyms, or visual cues to help remember key information more easily.
To find a pole's height, so neat, use sine for the rise, and make it complete.
Imagine a tree bending with the wind, its top touches the ground and we find its height with a trusty sine.
SOH-CAH-TOA helps remember sin, cos, and tan: Sine is Opposite over Hypotenuse, Cosine is Adjacent, and Tangent is Opposite over Adjacent.
Review key concepts with flashcards.
Review the Definitions for terms.
Term: Trigonometric Ratios
Definition:
Ratios derived from the angles of a right-angled triangle, specifically sine, cosine, and tangent.
Term: Angle of Elevation
Definition:
The angle formed by the line of sight and the horizontal when looking up at an object.
Term: Angle of Depression
Definition:
The angle formed by the line of sight and the horizontal when looking down at an object.