Angles Between Altitudes Of Triangle ABC A Comprehensive Solution

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Hey guys! Ever found yourself scratching your head over the angles formed by the altitudes of a triangle? Well, you're not alone! This topic can seem a bit daunting at first, but trust me, we'll break it down together and make it super clear. In this guide, we'll dive deep into how to find these angles, using a specific triangle ABC with vertices A(-4, -2), B(-3, 3), and C(5, -4) as our example. Get ready to sharpen those geometry skills!

Understanding Altitudes and Their Significance

Before we jump into calculations, let's make sure we're all on the same page about what an altitude actually is. In geometry, an altitude of a triangle is a line segment drawn from a vertex perpendicular to the opposite side (or the extension of the opposite side). Every triangle has three altitudes, one from each vertex. These altitudes play a crucial role in various geometric properties and calculations, including finding the area of a triangle and understanding its orthocenter (the point where all three altitudes intersect).

Why are altitudes so important, you ask? Well, for starters, they provide a direct measure of the "height" of a triangle relative to a specific base. Remember the formula for the area of a triangle: 1/2 * base * height? That "height" is precisely the length of the altitude! Moreover, the altitudes and their intersection point, the orthocenter, reveal a lot about the triangle's shape and symmetry. Acute triangles have their orthocenter inside the triangle, right triangles have it at the vertex where the right angle is formed, and obtuse triangles have it outside the triangle. Understanding these relationships is key to tackling more complex geometric problems.

Now, let's think about how the angles between these altitudes are related to the original triangle's angles. This is where things get interesting! The angles formed by the altitudes are closely connected to the angles of the triangle itself. In fact, they are supplementary to the original angles, meaning they add up to 180 degrees. This relationship is a fundamental concept we'll use to solve our problem. So, keep this in mind as we move forward: finding the angles between the altitudes ultimately boils down to understanding the angles within the original triangle.

Step-by-Step Guide to Finding the Angles

Alright, let's get our hands dirty and start calculating! We'll use the given vertices A(-4, -2), B(-3, 3), and C(5, -4) to find the angles between the altitudes of triangle ABC. We'll break this down into manageable steps, making it super easy to follow along. Here's the plan:

  1. Find the Slopes of the Sides: We'll start by calculating the slopes of the three sides of the triangle: AB, BC, and CA. Remember, the slope of a line is a measure of its steepness and is calculated as (change in y) / (change in x). This is a crucial first step because the slope of a line is directly related to the slope of a line perpendicular to it.
  2. Determine the Slopes of the Altitudes: Since altitudes are perpendicular to the sides, we'll use the fact that the product of the slopes of two perpendicular lines is -1. This will allow us to easily find the slopes of the three altitudes.
  3. Find the Equations of the Altitudes: With the slopes and a point on each altitude (the vertex from which it's drawn), we can write the equations of the lines representing the altitudes. We'll use the point-slope form of a line (y - y1 = m(x - x1)) for this.
  4. Calculate the Angles Between the Altitudes: Finally, we'll use the slopes of the altitudes to find the angles between them. We'll use the formula for the tangent of the angle between two lines: tan(θ) = |(m1 - m2) / (1 + m1 * m2)|, where m1 and m2 are the slopes of the lines. Once we have the tangent, we can use the arctangent function (tan^-1) to find the angle itself.

Let's get started with the first step!

1. Calculating the Slopes of the Sides

To kick things off, we need to find the slopes of the sides of our triangle ABC. Remember, the slope (m) between two points (x1, y1) and (x2, y2) is calculated as:

m = (y2 - y1) / (x2 - x1)

Let's apply this to our sides:

  • Slope of AB (mAB): Using points A(-4, -2) and B(-3, 3), we get: mAB = (3 - (-2)) / (-3 - (-4)) = 5 / 1 = 5
  • Slope of BC (mBC): Using points B(-3, 3) and C(5, -4), we get: mBC = (-4 - 3) / (5 - (-3)) = -7 / 8
  • Slope of CA (mCA): Using points C(5, -4) and A(-4, -2), we get: mCA = (-2 - (-4)) / (-4 - 5) = 2 / -9 = -2/9

So, we've found the slopes of the sides: mAB = 5, mBC = -7/8, and mCA = -2/9. This is a crucial step because these slopes will help us determine the slopes of the altitudes, which are perpendicular to these sides. Keep these values handy – we'll need them in the next step!

2. Finding the Slopes of the Altitudes

Now that we have the slopes of the sides, let's find the slopes of the altitudes. Remember, altitudes are perpendicular to the sides, and the product of the slopes of two perpendicular lines is -1. This gives us a neat little shortcut to calculate the altitude slopes.

  • Altitude from C to AB: Let's call the slope of this altitude mC. Since it's perpendicular to AB (slope mAB = 5), we have: mC * 5 = -1 mC = -1/5
  • Altitude from A to BC: Let's call the slope of this altitude mA. Since it's perpendicular to BC (slope mBC = -7/8), we have: mA * (-7/8) = -1 mA = 8/7
  • Altitude from B to CA: Let's call the slope of this altitude mB. Since it's perpendicular to CA (slope mCA = -2/9), we have: mB * (-2/9) = -1 mB = 9/2

Great! We've now calculated the slopes of the altitudes: mC = -1/5, mA = 8/7, and mB = 9/2. These values are essential for finding the equations of the altitude lines, which we'll tackle in the next step. You're doing awesome – keep up the momentum!

3. Determining the Equations of the Altitudes

With the slopes of the altitudes in hand, we can now find their equations. We'll use the point-slope form of a line, which is a super handy way to write the equation of a line when you know its slope (m) and a point (x1, y1) it passes through:

y - y1 = m(x - x1)

Let's find the equations for each altitude:

  • Altitude from C to AB: This altitude passes through point C(5, -4) and has a slope of mC = -1/5. Plugging these values into the point-slope form, we get: y - (-4) = (-1/5)(x - 5) y + 4 = (-1/5)x + 1 y = (-1/5)x - 3 (Equation of altitude from C)
  • Altitude from A to BC: This altitude passes through point A(-4, -2) and has a slope of mA = 8/7. Plugging these values into the point-slope form, we get: y - (-2) = (8/7)(x - (-4)) y + 2 = (8/7)x + 32/7 y = (8/7)x + 18/7 (Equation of altitude from A)
  • Altitude from B to CA: This altitude passes through point B(-3, 3) and has a slope of mB = 9/2. Plugging these values into the point-slope form, we get: y - 3 = (9/2)(x - (-3)) y - 3 = (9/2)x + 27/2 y = (9/2)x + 33/2 (Equation of altitude from B)

Fantastic! We've successfully found the equations of all three altitudes. Now, we're just one step away from finding the angles between them. Get ready to put those slopes to work in the final calculation!

4. Calculating the Angles Between the Altitudes

Alright, we've reached the final showdown! We're going to use the slopes of the altitudes we calculated earlier to find the angles between them. The magic formula we'll use is:

tan(θ) = |(m1 - m2) / (1 + m1 * m2)|

Where θ is the angle between the two lines, and m1 and m2 are their slopes. Remember, we're dealing with absolute values here, so we'll always get a positive value for the tangent. This will give us the acute angle between the lines, and if we need the obtuse angle, we can simply subtract the acute angle from 180 degrees.

Let's calculate the angles:

  • Angle between altitudes from C and A: Slopes are mC = -1/5 and mA = 8/7. Plugging these into our formula: tan(θ) = |((-1/5) - (8/7)) / (1 + (-1/5) * (8/7))| tan(θ) = |(-7/35 - 40/35) / (1 - 8/35)| tan(θ) = |(-47/35) / (27/35)| tan(θ) = |-47/27| tan(θ) = 47/27 θ = arctan(47/27) ≈ 60.1 degrees The supplementary angle is 180 - 60.1 = 119.9 degrees.

  • Angle between altitudes from C and B: Slopes are mC = -1/5 and mB = 9/2. Plugging these into our formula: tan(θ) = |((-1/5) - (9/2)) / (1 + (-1/5) * (9/2))| tan(θ) = |(-2/10 - 45/10) / (1 - 9/10)| tan(θ) = |(-47/10) / (1/10)| tan(θ) = |-47| tan(θ) = 47 θ = arctan(47) ≈ 88.8 degrees The supplementary angle is 180 - 88.8 = 91.2 degrees.

  • Angle between altitudes from A and B: Slopes are mA = 8/7 and mB = 9/2. Plugging these into our formula: tan(θ) = |((8/7) - (9/2)) / (1 + (8/7) * (9/2))| tan(θ) = |(16/14 - 63/14) / (1 + 72/14)| tan(θ) = |(-47/14) / (86/14)| tan(θ) = |-47/86| tan(θ) = 47/86 θ = arctan(47/86) ≈ 28.7 degrees The supplementary angle is 180 - 28.7 = 151.3 degrees.

And there you have it! We've successfully calculated the angles between the altitudes of triangle ABC. Remember, these angles are closely related to the angles of the original triangle, and understanding this relationship is key to solving geometry problems. You've done a fantastic job following along, guys! Geometry might seem tricky at times, but with a step-by-step approach and a little practice, you can conquer any problem. Keep up the great work!

Key Takeaways and Further Exploration

Let's quickly recap what we've learned and think about where you can go next with this knowledge. We've covered:

  • The definition of an altitude and its importance in triangle geometry.
  • The relationship between the slopes of perpendicular lines and how to use it to find the slopes of altitudes.
  • The point-slope form of a line and how to use it to find the equations of altitudes.
  • The formula for the tangent of the angle between two lines and how to use it to calculate the angles between the altitudes.
  • The supplementary relationship between the angles formed by the altitudes and the original angles of the triangle.

Now that you've mastered this, you can explore related concepts like:

  • The orthocenter of a triangle: The point where all three altitudes intersect. Try finding the coordinates of the orthocenter for triangle ABC!
  • The relationship between the angles of the original triangle and the angles formed by the feet of the altitudes: This leads to some interesting geometric properties and theorems.
  • Applications of altitudes in real-world scenarios: Think about how altitudes can be used in surveying, architecture, and engineering.

Geometry is a fascinating field with endless possibilities for exploration. Keep practicing, keep asking questions, and most importantly, keep having fun with it! You've got this, guys!