Trigonometry Problem: Finding Tan(ACB) In Triangle ABC
Hey guys! Let's dive into an interesting trigonometry problem today. We're going to tackle a triangle question that involves finding the tangent of an angle given some side lengths and a relationship between two angles. So, buckle up, and let's get started!
Understanding the Problem
Before we jump into solving, let's make sure we fully grasp the problem statement. We're given a triangle ABC where the measure of angle ABC is twice the measure of angle ACB. We also know the lengths of sides AB and BC, which are 10 units and 22 units, respectively. Our mission is to find the value of tan(ACB). Sounds like a fun challenge, right?
To solve this, we'll need to use our knowledge of trigonometry, triangle properties, and maybe a little bit of algebraic manipulation. Don't worry, we'll break it down step by step so it's super clear.
Setting up the Triangle and Key Information
First things first, let's visualize the triangle. Imagine triangle ABC, where:
- |AB| = 10 units
- |BC| = 22 units
- m(ABC) = 2 * m(ACB)
Let's denote the measure of angle ACB as θ (theta). This means the measure of angle ABC is 2θ. Now, we need to find tan(θ). This is where the fun begins!
We know that the tangent of an angle in a right triangle is the ratio of the opposite side to the adjacent side. However, we don't know if our triangle ABC is a right triangle. So, we'll need to use some clever tricks and trigonometric identities to solve this problem.
Strategic Approaches to the Solution
Okay, so how do we tackle this? One common strategy in geometry problems is to look for ways to create right triangles. Right triangles are our best friends in trigonometry because they allow us to use the basic trigonometric ratios (sine, cosine, tangent) directly.
The Angle Bisector Theorem
One approach we can try involves using the angle bisector theorem. Since we have a relationship between angles ABC and ACB, we might be able to create some similar triangles by bisecting angle ABC. Let's draw a line from B that bisects angle ABC, and let's call the point where this line intersects side AC as D. So, BD bisects angle ABC, meaning angles ABD and DBC are both equal to θ.
Now, we have two new triangles: ABD and DBC. This is a good start! The angle bisector theorem tells us that the ratio of the sides adjacent to the bisected angle is equal to the ratio of the segments created on the opposite side. In our case:
AB / BC = AD / DC
Plugging in the given values:
10 / 22 = AD / DC
Simplifying the fraction:
5 / 11 = AD / DC
Let's say AD = 5x and DC = 11x. This means AC = AD + DC = 5x + 11x = 16x.
Law of Sines and the Double Angle Formula
Another powerful tool in our arsenal is 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 ABC, this means:
AB / sin(ACB) = BC / sin(BAC) = AC / sin(ABC)
Let's plug in what we know:
10 / sin(θ) = 22 / sin(BAC) = 16x / sin(2θ)
Now, we can use the double angle formula for sine, which states that sin(2θ) = 2 * sin(θ) * cos(θ). Substituting this into our equation:
10 / sin(θ) = 16x / (2 * sin(θ) * cos(θ))
We can simplify this equation by canceling out sin(θ) from both denominators:
10 = 8x / cos(θ)
Solving for cos(θ):
cos(θ) = 8x / 10 = 4x / 5
This is progress! We have an expression for cos(θ) in terms of x.
Solving for tan(θ)
Alright, we're getting closer to our goal. We need to find tan(θ), and we have an expression for cos(θ). Remember that tan(θ) = sin(θ) / cos(θ). So, if we can find sin(θ) in terms of x, we'll be golden.
Using the Pythagorean Identity
We can use the Pythagorean identity, which states that sin²(θ) + cos²(θ) = 1. We already have cos(θ) = 4x / 5, so let's substitute that into the identity:
sin²(θ) + (4x / 5)² = 1
sin²(θ) + 16x² / 25 = 1
Solving for sin²(θ):
sin²(θ) = 1 - 16x² / 25
sin²(θ) = (25 - 16x²) / 25
Taking the square root of both sides:
sin(θ) = √((25 - 16x²) / 25)
sin(θ) = √(25 - 16x²) / 5
Now we have sin(θ) in terms of x as well!
Calculating tan(θ)
Finally, we can calculate tan(θ) using the formula tan(θ) = sin(θ) / cos(θ):
tan(θ) = (√(25 - 16x²) / 5) / (4x / 5)
We can simplify this by canceling out the 5s:
tan(θ) = √(25 - 16x²) / 4x
Finding the Value of x
Okay, we have tan(θ) in terms of x, but we need a numerical value. To find x, let's go back to the Law of Sines equation:
10 / sin(θ) = 22 / sin(BAC)
We also know that the sum of angles in a triangle is 180 degrees, so:
θ + 2θ + BAC = 180
3θ + BAC = 180
BAC = 180 - 3θ
So, sin(BAC) = sin(180 - 3θ) = sin(3θ). We can use the triple angle formula for sine:
sin(3θ) = 3sin(θ) - 4sin³(θ)
Now we can rewrite our Law of Sines equation:
10 / sin(θ) = 22 / (3sin(θ) - 4sin³(θ))
Divide both sides by sin(θ):
10 = 22 / (3 - 4sin²(θ))
Substitute sin²(θ) = (25 - 16x²) / 25:
10 = 22 / (3 - 4((25 - 16x²) / 25))
Now, let's solve for x. This involves some algebraic manipulation:
10(3 - 4((25 - 16x²) / 25)) = 22
3 - 4((25 - 16x²) / 25) = 2.2
3 - (100 - 64x²) / 25 = 2.2
75 - (100 - 64x²) = 55
75 - 100 + 64x² = 55
64x² = 80
x² = 80 / 64 = 5 / 4
x = √(5 / 4) = √5 / 2
Final Calculation of tan(θ)
We finally have the value of x! Now we can plug it back into our equation for tan(θ):
tan(θ) = √(25 - 16x²) / 4x
tan(θ) = √(25 - 16(5 / 4)) / (4 * (√5 / 2))
tan(θ) = √(25 - 20) / (2√5)
tan(θ) = √5 / (2√5)
tan(θ) = 1 / 2
Conclusion
Woohoo! We did it! The value of tan(ACB) is 1/2. That was quite a journey, but we made it through using a combination of trigonometric identities, the Law of Sines, the angle bisector theorem, and some algebraic problem-solving skills.
Remember, the key to tackling these kinds of problems is to break them down into smaller, manageable steps. Look for opportunities to create right triangles, use trigonometric identities, and apply theorems like the Law of Sines and the angle bisector theorem. Keep practicing, and you'll become a trigonometry pro in no time!
If you guys have any questions or want to explore more problems like this, let me know in the comments. Keep up the great work, and I'll see you in the next problem!