Triangles: Sides 6, 8, Angle 30° - How Many?

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Hey guys! Let's dive into a cool geometry problem today. We're going to figure out how many different triangles we can draw if we know two sides and one angle. Specifically, we're looking at triangles with sides of length 6 and 8 units, and one angle that measures 30 degrees. Sounds like a fun puzzle, right? So grab your thinking caps, and let's get started!

Understanding the Triangle Problem

Before we jump into solving this, let's make sure we understand the basics. When we talk about constructing a triangle, we need to know certain information, like the lengths of the sides and the measures of the angles. In our case, we know two sides (6 and 8 units) and one angle (30 degrees). The big question is, does this information uniquely define a triangle? Or could we possibly draw more than one triangle with these measurements?

When you're tackling geometry problems, it's super useful to visualize what's going on. Try sketching out a few different triangles that fit the description. This can help you get a feel for the problem and might even give you some clues about the solution. What happens if you try drawing the 30-degree angle between the two given sides? What if the 30-degree angle is opposite one of the sides? Exploring these different scenarios is key to cracking this problem.

The crucial concept here is the Side-Side-Angle (SSA) condition for triangle congruence. Unlike Side-Angle-Side (SAS) or Angle-Side-Angle (ASA), SSA doesn't guarantee a unique triangle. This means that knowing two sides and an angle (not included between the sides) might lead to zero, one, or even two possible triangles. This is exactly why our problem is so interesting! We need to carefully consider the different possibilities to determine the correct answer.

Breaking Down the Possibilities

Okay, so let's systematically break down the different ways we can construct this triangle. We have two sides, let's call them a = 6 and b = 8, and an angle, let's call it α = 30°. Remember, this is an SSA situation, so we need to be extra careful.

  • Scenario 1: The 30° angle is opposite the side with length 6. Imagine side b (length 8) as a base. Now, we need to draw side a (length 6) opposite the 30° angle. Think of side a as a swinging arm, pivoted at one end of side b. The question is, how many times can this arm intersect the line that forms the other side of the 30° angle? This will tell us how many triangles we can form.

  • Scenario 2: The 30° angle is opposite the side with length 8. Now, let's flip things around. Imagine side a (length 6) as the base. Side b (length 8) is now opposite the 30° angle. Again, we can think of side b as a swinging arm. How many times can this longer arm intersect the line that forms the other side of the 30° angle? This situation might behave differently from the first scenario.

By carefully analyzing these two scenarios, we can start to narrow down the possible number of triangles. This is where our understanding of trigonometry, particularly the Law of Sines, can come in handy. But before we get too deep into the calculations, let's keep visualizing and sketching. A good diagram can often save you a ton of time and effort!

Using the Law of Sines

The Law of Sines is our secret weapon for this problem! It tells us the relationship between the sides of a triangle and the sines of their opposite angles. In simple terms, it states that for any triangle ABC:

a / sin(A) = b / sin(B) = c / sin(C)

Where a, b, and c are the side lengths, and A, B, and C are the angles opposite those sides.

In our case, we know a = 6, b = 8, and α = 30°. Let's use the Law of Sines to figure out the possible values for angle B, which is opposite side b:

6 / sin(30°) = 8 / sin(B)

Since sin(30°) = 0.5, we can simplify the equation:

6 / 0.5 = 8 / sin(B)
12 = 8 / sin(B)
sin(B) = 8 / 12 = 2 / 3

Now, we need to find the angle B whose sine is 2/3. Remember, the sine function has values between -1 and 1, and there can be two angles in the range of 0° to 180° that have the same sine value. This is a critical point! It means we might have two possible values for angle B.

Let's call the first angle B₁ = arcsin(2/3). This is the principal value we get from the inverse sine function. To find the second possible angle, B₂, we use the fact that sin(180° - x) = sin(x). So:

B₂ = 180° - arcsin(2/3)

Now we have two potential angles for B. But does this mean we can form two triangles? Not necessarily! We need to check if these angles make sense in the context of a triangle. Remember, the sum of the angles in a triangle must be 180°.

Checking for Valid Triangles

We've found two possible angles for B, let's call them B₁ and B₂. To see if they actually form valid triangles, we need to check if the sum of angles α (30°), B₁, and the remaining angle C₁ is less than or equal to 180°, and the sum of angles α (30°), B₂, and the remaining angle C₂ is less than or equal to 180°.

Let's think about what this means. If α + B₁ is less than 180°, then we can find a valid angle C₁ by subtracting their sum from 180°. The same goes for α + B₂. However, if either α + B₁ or α + B₂ is greater than 180°, then we can't form a triangle because we'd end up with a negative or zero angle for C, which is impossible.

So, we need to calculate B₁ and B₂ (using a calculator or trigonometric tables) and then check these sums. This step is essential to avoid counting invalid triangles.

Let's say we calculate arcsin(2/3) and find that B₁ is approximately 41.8°. Then, B₂ would be approximately 180° - 41.8° = 138.2°.

Now let's check the sums:

  • α + B₁ = 30° + 41.8° = 71.8° (which is less than 180°)
  • α + B₂ = 30° + 138.2° = 168.2° (which is also less than 180°)

Since both sums are less than 180°, it seems like we can form two triangles! But before we confidently declare our answer, there's one more thing we need to consider.

The Ambiguous Case of SSA

Remember we talked about the ambiguous case of the Side-Side-Angle (SSA) condition? This is where things can get a little tricky. Even if we find two possible angles for B that satisfy the Law of Sines and result in valid angle sums, we still need to make sure that the sides we're given can actually form a triangle with those angles.

Think back to our "swinging arm" analogy. We imagined side a as an arm swinging from one end of side b. We found two possible intersection points, which gave us our two possible angles for B. However, it's possible that one of those intersection points is actually behind the starting point of the arm, meaning it wouldn't form a valid triangle.

To check for this, we can use the triangle inequality theorem. This theorem states that the sum of any two sides of a triangle must be greater than the third side. We need to make sure this holds true for both of our potential triangles.

For each possible triangle (with angles α, B₁, C₁ and α, B₂, C₂), we would calculate the remaining side length using the Law of Sines or the Law of Cosines. Then, we'd check if the triangle inequality holds. If it doesn't, that triangle is not possible, and we need to discard it.

This final check is super important to ensure we have the correct answer. Sometimes, even though we've done all the math right, the geometry just doesn't work out for one of the solutions!

The Final Answer

Alright guys, after carefully considering all the possibilities, using the Law of Sines, checking the angle sums, and applying the triangle inequality theorem, we can finally arrive at the answer! Drumroll, please...

In this specific case, with sides of lengths 6 and 8 units and a 30-degree angle, it turns out that we can draw two different triangles. This is because the SSA condition allows for the possibility of multiple solutions, and both potential triangles we found are geometrically valid.

So, the correct answer is B) 2.

Key Takeaways

This problem was a fantastic exercise in understanding triangle geometry and the ambiguous case of SSA. Here are the key things we learned:

  • The Side-Side-Angle (SSA) condition doesn't guarantee a unique triangle.
  • The Law of Sines is a powerful tool for solving triangle problems, but it can sometimes lead to multiple solutions.
  • We must always check if the solutions we find are geometrically valid by considering angle sums and the triangle inequality theorem.
  • Visualizing the problem and sketching diagrams can be incredibly helpful.

Geometry problems like this can seem tricky at first, but by breaking them down step by step and applying the right concepts, you can conquer them! Keep practicing, and you'll become a triangle-solving master in no time.