Triangle ABC Side Ratio Calculation CosA=4/5 And TanB=2/3

by Admin 58 views

Hey there, math enthusiasts! Today, we're diving deep into a fascinating geometry problem involving triangle ABC. We're given that angles A and B are acute, with cosA = 4/5 and tanB = 2/3. Our mission? To find the holy grail the ratio AB : BC : CA. Buckle up, because we're about to embark on a mathematical journey filled with trigonometric identities, the Law of Sines, and a dash of clever problem-solving!

Cracking the CosA and TanB Clues

So, we're presented with the information that cosA = 4/5 and tanB = 2/3. These values are our starting points, the breadcrumbs that will lead us to the solution. But what do they really tell us about the triangle? Let's break it down.

When we see cosA = 4/5, we should immediately think of the SOH CAH TOA mnemonic. Cosine is Adjacent over Hypotenuse. This means in a right-angled triangle (which we can hypothetically create as a reference), the side adjacent to angle A is 4 units, and the hypotenuse is 5 units. Using the Pythagorean theorem, we can find the opposite side: sqrt(5^2 - 4^2) = 3. Therefore, we now know all the trigonometric ratios for angle A: cosA = 4/5, sinA = 3/5, and tanA = 3/4. These values become our foundational pillars.

Similarly, tanB = 2/3 gives us a ratio of Opposite to Adjacent. Imagine another right-angled triangle where the side opposite angle B is 2 units and the adjacent side is 3 units. Again, using the Pythagorean theorem, the hypotenuse is sqrt(2^2 + 3^2) = sqrt(13). Thus, we can find sinB = 2/sqrt(13) and cosB = 3/sqrt(13). Guys, recognizing these relationships is absolutely key to unlocking the rest of the problem. These values are crucial for moving forward, acting like essential pieces in our puzzle.

Remember, trigonometric ratios are like secret codes that reveal the relationships between angles and sides in triangles. Mastering these ratios is like gaining fluency in the language of triangles! Now that we have cracked the codes for angles A and B, we're ready to move on to the next stage of our mathematical adventure.

The Law of Sines to the Rescue

Now that we have sinA and sinB, we need a powerful tool to connect these values to the sides of our triangle ABC. Enter the Law of Sines! This theorem is like a bridge that links angles and their opposite sides. It states that in any triangle, the ratio of a side length to the sine of its opposite angle is constant. Mathematically, it's expressed as:

a / sinA = b / sinB = c / sinC

Where a, b, and c are the side lengths, and A, B, and C are the angles opposite those sides, respectively. In our case, we want to find the ratio AB : BC : CA, which corresponds to c : a : b. So, the Law of Sines is perfectly suited to our task. This powerful theorem is our best friend in solving triangles.

We already have sinA = 3/5 and sinB = 2/sqrt(13). To use the Law of Sines effectively, we also need to find sinC. Remember that the angles in a triangle add up to 180 degrees (or π radians). Therefore, C = 180° - A - B. Finding the sine of C might seem daunting, but here's where our trigonometric identities come to the rescue again! We can use the sine addition formula: sin(A + B) = sinA cosB + cosA sinB. This handy formula will help us find the sine of the sum of angles A and B, which is crucial in determining sinC. This is a key step that allows us to connect the angles and sides of the triangle using a well-established trigonometric relationship.

So, let's calculate sin(A + B). We know sinA = 3/5, cosA = 4/5, sinB = 2/sqrt(13), and cosB = 3/sqrt(13). Plugging these values into the formula, we get:

sin(A + B) = (3/5)(3/sqrt(13)) + (4/5)(2/sqrt(13)) = (9/5sqrt(13)) + (8/5sqrt(13)) = 17/5sqrt(13)

Since C = 180° - (A + B), we have sinC = sin(180° - (A + B)) = sin(A + B) = 17/5sqrt(13). Now, we've got all the pieces we need to apply the Law of Sines and unlock the ratio of the sides. Remember, each calculated piece fits perfectly into the puzzle we are building.

Unveiling the Ratio AB : BC : CA

Alright, we've done the groundwork, laid the foundation, and now it's time for the grand finale finding the ratio AB : BC : CA. We've got our ingredients sinA = 3/5, sinB = 2/sqrt(13), and sinC = 17/5sqrt(13). Let's plug these values into the Law of Sines:

a / sinA = b / sinB = c / sinC

Let's rewrite this in terms of the sides we want to find:

a / (3/5) = b / (2/sqrt(13)) = c / (17/5sqrt(13))

To make things easier, let's find a common multiplier, k, such that:

a = (3/5)k b = (2/sqrt(13))k c = (17/5sqrt(13))k

We want to find the ratio a : b : c, so we can simply divide each side by k and then find a common denominator to get rid of the fractions. We have:

a : b : c = (3/5) : (2/sqrt(13)) : (17/5sqrt(13))

To eliminate the fractions, we can multiply each term by 5sqrt(13). This gives us:

a : b : c = 3sqrt(13) : 10 : 17

So, there we have it! The ratio AB : BC : CA, which is the same as c : a : b, is:

AB : BC : CA = 17 : 3sqrt(13) : 10

We've successfully navigated the trigonometric terrain and emerged victorious with our ratio! Each step, from understanding the initial clues to applying the Law of Sines, was crucial in reaching our destination. Great job, guys!

Visualizing the Solution

Sometimes, a visual aid can really solidify our understanding. Imagine triangle ABC with the given conditions. Angle A is acute, and angle B is also acute. We've determined the ratio of the sides, but let's think about what this means geometrically.

Since cosA = 4/5, angle A is relatively small. Since tanB = 2/3, angle B is also moderately sized. Angle C, being the remaining angle, is determined by the sum of angles in a triangle. The ratio we found tells us that side AB (opposite angle C) is the longest, followed by side BC (opposite angle A), and then side CA (opposite angle B).

Creating a rough sketch of the triangle, even if it's not perfectly to scale, can help us see the relationships between the sides and angles. This visual confirmation reinforces our understanding and adds another layer to our comprehension. Visualizing the problem can sometimes reveal insights that might be missed when looking purely at the equations. It's like seeing the landscape after studying the map; it all starts to make sense in a more intuitive way.

Moreover, visualizing the solution allows us to double-check if our answer makes sense in the context of the problem. Do the side lengths and angles seem proportional to each other? Does the triangle look reasonable given the initial conditions? These are important questions to ask ourselves to ensure we haven't made any major errors along the way.

Key Takeaways and Practice

Wow, what a journey! We've successfully cracked this triangle problem by leveraging our understanding of trigonometric ratios, the Law of Sines, and some clever problem-solving techniques. Let's recap the key takeaways from our adventure:

  • Trigonometric Ratios are Your Friends: cosA = 4/5 and tanB = 2/3 were our starting points. Knowing these ratios allowed us to find sinA, cosB, and sinB, which were essential for applying the Law of Sines.
  • Law of Sines is a Powerful Tool: This theorem connects the sides and angles of any triangle, making it invaluable for solving problems like this one.
  • Trigonometric Identities are Your Allies: The sine addition formula helped us find sinC, which was crucial for completing our Law of Sines equation.
  • Visualize the Problem: Drawing a diagram can provide valuable insights and help you understand the relationships between the sides and angles.

To truly master these concepts, practice is key! Try solving similar problems with different given values. Experiment with different triangles and explore the relationships between sides and angles. The more you practice, the more comfortable you'll become with these techniques.

And remember, mathematics is not just about finding the right answer; it's about the journey of exploration and discovery. Each problem is a puzzle waiting to be solved, and each solution is a step forward in your mathematical understanding. So, keep exploring, keep questioning, and keep practicing! You've got this!

Original Question

In triangle ABC, angles A and B are acute where cosA equals 4/5 and tanB equals 2/3. What is the ratio AB : BC : CA?