Finding Tan Α In Equilateral Triangle ABC: A Step-by-Step Guide

Oct 15, 2025 by ADMIN 64 views

Hey guys! Today, we're diving into a cool geometry problem involving an equilateral triangle and finding the tangent of an angle. It might seem tricky at first, but don't worry, we'll break it down step-by-step so it's super easy to follow. Let's get started!

Understanding the Problem: Equilateral Triangle ABC

So, the problem throws us an equilateral triangle ABC. Remember what that means? An equilateral triangle is special because all its sides are equal in length, and all its angles are equal too – each measuring 60 degrees. This is our starting point, and these facts are crucial for solving the problem. Now, we've got a line segment drawn from vertex A, but it doesn't bisect the angle perfectly. This creates an angle, alpha (α), which is what we're ultimately trying to figure out the tangent of.

The given equation, $\frac{\frac{\sqrt{3}}{2}a}{\frac{a}{6}} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{6}} = \frac{\sqrt{3}}{2} \times 6 = 3\sqrt{3}$ , might look a bit intimidating, but it's actually helping us understand the ratios within the triangle. Let's break down what this equation tells us. The expression $\frac{\sqrt{3}}{2}a$ likely represents the altitude (height) of the equilateral triangle. Why? Because in an equilateral triangle, the altitude bisects the base and forms two 30-60-90 right triangles. The ratio of sides in a 30-60-90 triangle is a well-known concept in geometry, and the altitude corresponds to the side opposite the 60-degree angle, which has a length of $\frac{\sqrt{3}}{2}$ times the side length of the equilateral triangle (which we're calling 'a').

The term $\frac{a}{6}$ likely refers to a segment of the base BC. The equation then simplifies to $3\sqrt{3}$ , which represents a ratio related to the triangle's dimensions and will probably be useful later when we start calculating $tan \alpha$ . The information $5|BD| = |DC|$ tells us that the side BC is divided into two segments, BD and DC, where DC is five times longer than BD. This is a crucial piece of information for figuring out the lengths of these segments relative to the side length of the triangle. Knowing the proportions of BD and DC will allow us to use trigonometric relationships to find $tan \alpha$ .

Key Properties of Equilateral Triangles

Before we move on, let's quickly recap the vital properties of equilateral triangles, as they're the foundation for solving this problem:

All three sides are equal in length.
All three angles are equal, each measuring 60 degrees.
The altitude (height) bisects the base and the vertex angle.
The altitude also acts as the median and the angle bisector.
It can be divided into two 30-60-90 right triangles, which have special side ratios.

Keeping these properties in mind will help us visualize the problem and apply the correct trigonometric principles.

Setting Up the Problem for Trigonometry

Okay, let's translate the geometric information into something we can work with trigonometrically. We're given that 5|BD| = |DC|. This means that the segment DC is five times the length of segment BD. Let's assume the side length of our equilateral triangle ABC is 'a'. This will help us express the lengths of BD and DC in terms of 'a'.

If we let the length of BD be 'x', then the length of DC is '5x'. Since BD and DC together make up the base BC of the equilateral triangle, we know that BD + DC = BC. Substituting our expressions, we get x + 5x = a, which simplifies to 6x = a. Therefore, x = a/6, meaning |BD| = a/6 and |DC| = 5a/6. Now we have the lengths of BD and DC in terms of the side length 'a' of the equilateral triangle. This is a crucial step because it allows us to relate these lengths to the overall dimensions of the triangle.

Visualizing the Triangle and the Angle α

Imagine drawing the equilateral triangle ABC. Now, picture the line segment AD, which creates the angle α (∠BAD) that we're interested in. This line segment AD divides the triangle into two smaller triangles: triangle ABD and triangle ADC. Our goal is to find tan α, which is the ratio of the opposite side to the adjacent side in a right triangle. However, triangle ABD isn't necessarily a right triangle, so we'll need to use some clever trigonometry and the information we've gathered to find this ratio.

To find tan α, we'll likely need to use the Law of Sines or the Law of Cosines, or possibly break down the triangle further by drawing an altitude. We need to find a way to relate α to the known side lengths and angles within the triangle. Remember that the angles of an equilateral triangle are all 60 degrees. This gives us a starting point. We know ∠ABC = 60 degrees and we've expressed BD and DC in terms of 'a'. Now we need to strategize how to use this information to isolate tan α.

Applying Trigonometric Principles to Find tan α

Now comes the fun part – applying our trigonometric knowledge to crack this problem! We've got the side lengths BD and DC in terms of 'a', and we know all the angles in the equilateral triangle are 60 degrees. Our mission is to find tan α. One approach we can use here involves the Law of Sines. The Law of Sines states that in any triangle, the ratio of the length of a side to the sine of its opposite angle is constant. In triangle ABD, this means:

$\frac{BD}{sin(\alpha)} = \frac{AD}{sin(60^{\circ})}$

And in triangle ADC, we have:

$\frac{DC}{sin(60^{\circ} - \alpha)} = \frac{AD}{sin(60^{\circ})}$

Notice that both equations have AD and sin(60°) in them. This is great because we can set them equal to each other, eliminating AD from the equation. By equating the two expressions for AD, we get:

$\frac{BD}{sin(\alpha)} = \frac{DC}{sin(60^{\circ} - \alpha)}$

Now, we can substitute the values we found earlier for BD (a/6) and DC (5a/6). This gives us:

$\frac{a/6}{sin(\alpha)} = \frac{5a/6}{sin(60^{\circ} - \alpha)}$

The 'a/6' terms cancel out, simplifying the equation to:

$\frac{1}{sin(\alpha)} = \frac{5}{sin(60^{\circ} - \alpha)}$

This equation relates α to a known angle (60 degrees). Now we need to use trigonometric identities to expand sin(60° - α) and solve for tan α.

Using the Sine Subtraction Identity

To proceed, we'll use the sine subtraction identity, which states:

sin(A - B) = sin(A)cos(B) - cos(A)sin(B)

Applying this to sin(60° - α), we get:

sin(60° - α) = sin(60°)cos(α) - cos(60°)sin(α)

We know that sin(60°) = $\frac{\sqrt{3}}{2}$ and cos(60°) = $\frac{1}{2}$ . Substituting these values, we get:

sin(60° - α) = $\frac{\sqrt{3}}{2}$ cos(α) - $\frac{1}{2}$ sin(α)

Now we can substitute this back into our equation:

$\frac{1}{sin(\alpha)} = \frac{5}{\frac{\sqrt{3}}{2}cos(\alpha) - \frac{1}{2}sin(\alpha)}$

Cross-multiplying gives us:

$\frac{\sqrt{3}}{2}$ cos(α) - $\frac{1}{2}$ sin(α) = 5sin(α)

Multiplying both sides by 2 to get rid of the fractions, we have:

$\sqrt{3}$ cos(α) - sin(α) = 10sin(α)

Isolating tan α and Finding the Solution

Now, let's isolate the terms with sine and cosine. Add sin(α) to both sides:

$\sqrt{3}$ cos(α) = 11sin(α)

To get tan α, we need to divide both sides by cos(α). Remember that tan(α) = sin(α)/cos(α). So, dividing both sides by cos(α) gives us:

$\sqrt{3}$ = 11 $\frac{sin(\alpha)}{cos(\alpha)}$

$\sqrt{3}$ = 11tan(α)

Finally, divide both sides by 11 to solve for tan α:

tan(α) = $\frac{\sqrt{3}}{11}$

And there you have it! The value of tan α in our equilateral triangle ABC is $\frac{\sqrt{3}}{11}$ .

Reviewing the Steps

Let's quickly recap the steps we took to solve this problem:

We understood the properties of equilateral triangles and the given information about the side ratios.
We expressed the lengths of BD and DC in terms of the side length 'a' of the triangle.
We applied the Law of Sines to triangles ABD and ADC.
We used the sine subtraction identity to expand sin(60° - α).
We simplified the equation and solved for tan α.

By breaking the problem down into smaller steps and using the appropriate trigonometric principles, we were able to find the solution. Geometry problems might seem tough at first, but with a systematic approach and a good understanding of trigonometric identities, you can conquer them!

Conclusion: The Beauty of Trigonometry

So, guys, we've successfully navigated through this geometry problem and found the value of tan α in an equilateral triangle. Remember, the key to solving these kinds of problems is to break them down into smaller, manageable steps. Use the given information wisely, apply the right trigonometric principles, and don't be afraid to use identities to simplify your equations.

Trigonometry is a powerful tool that helps us understand the relationships between angles and sides in triangles, and it's essential in many areas of math, science, and engineering. Keep practicing, and you'll become a geometry whiz in no time! Keep exploring, keep learning, and I'll catch you in the next one!