Trigonometry Problems: Finding Sin B And Tan X

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Hey guys! Let's dive into some trigonometry problems today. We'll be tackling questions involving right triangles and trigonometric functions like sine, cosine, and tangent. These are fundamental concepts in trigonometry, and mastering them will help you in various mathematical and real-world applications. So, grab your calculators and let's get started!

11. Finding sin B in a Right Triangle

Let's break down the first problem. We're given a right triangle ABC, where the right angle is at point A. We know that side AB is 3 cm and side BC is 6 cm. The task? Determine the value of sin B.

First things first, it's super important to visualize the triangle. Imagine a triangle where angle A is 90 degrees. AB is one of the legs, and BC is the hypotenuse (the side opposite the right angle). To find sin B, we need to remember the basic trigonometric ratios. Sine is defined as the ratio of the opposite side to the hypotenuse. In this case, the side opposite angle B is AC, and the hypotenuse is BC.

So, we have sin B = AC / BC. But wait, we don't know the length of AC! That's where the Pythagorean theorem comes in handy. This theorem states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Mathematically, it's expressed as a² + b² = c², where c is the hypotenuse.

In our triangle, this translates to AB² + AC² = BC². Plugging in the known values, we get 3² + AC² = 6². This simplifies to 9 + AC² = 36. Subtracting 9 from both sides gives us AC² = 27. Taking the square root of both sides, we find that AC = √27 = 3√3 cm. Now we have all the pieces of the puzzle!

We can now calculate sin B = AC / BC = (3√3) / 6. Simplifying this fraction, we get sin B = √3 / 2. So, the value of sin B in this triangle is √3 / 2. This is a classic example of how we use trigonometric ratios and the Pythagorean theorem together to solve problems involving right triangles. Remember, understanding the definitions of sine, cosine, and tangent and how they relate to the sides of a right triangle is key. And don't forget the Pythagorean theorem – it's your best friend when dealing with right triangles!

12. Determining tan X When cos X is Known

Next up, we have a problem where we're given cos X = 5/7, and we know that angle X is an acute angle (meaning it's between 0 and 90 degrees). The mission, should you choose to accept it, is to find the value of tan X. This problem highlights the relationship between different trigonometric functions.

To tackle this, remember the fundamental definitions: cosine (cos) is the ratio of the adjacent side to the hypotenuse, and tangent (tan) is the ratio of the opposite side to the adjacent side. So, if cos X = 5/7, we can imagine a right triangle where the side adjacent to angle X is 5 units long, and the hypotenuse is 7 units long.

Again, we need to use the Pythagorean theorem to find the length of the side opposite angle X. Let's call this side 'a'. According to the theorem, a² + 5² = 7². This simplifies to a² + 25 = 49. Subtracting 25 from both sides, we get a² = 24. Taking the square root, we find that a = √24 = 2√6. Now we know the length of the opposite side.

Now, let's calculate tan X. Remember, tan X = opposite / adjacent. So, tan X = (2√6) / 5. And there you have it! The value of tan X is (2√6) / 5. This problem beautifully illustrates how we can use the given information about one trigonometric function (in this case, cosine) to find the value of another (tangent). The key is to use the Pythagorean theorem to find the missing side length and then apply the definitions of the trigonometric ratios.

13. Solving for sin A in a Right Triangle

Let's tackle our third problem. We're told we have a right triangle ABC, and the right angle is at point C. We're also given that sin A = something (the value was missing in the original prompt, so let's assume sin A = 3/5 for the sake of this example). The goal is to find out more about the triangle, specifically using the given sin A value.

First off, remember what sine means. Sine of an angle is the ratio of the opposite side to the hypotenuse. So, if sin A = 3/5, this tells us that the side opposite angle A (which is BC) is 3 units long, and the hypotenuse (which is AB) is 5 units long. We can already start visualizing this triangle!

Now, let's use our trusty friend, the Pythagorean theorem, to find the length of the remaining side, AC. We know that AC² + BC² = AB². Plugging in the values we have, we get AC² + 3² = 5². This simplifies to AC² + 9 = 25. Subtracting 9 from both sides gives us AC² = 16. Taking the square root, we find that AC = 4 units.

Now that we know all three sides of the triangle, we can actually find the values of all the trigonometric functions for angles A and B (since C is the right angle). Let's do that! We already know sin A = 3/5.

  • cos A (adjacent / hypotenuse) = 4/5
  • tan A (opposite / adjacent) = 3/4

Now, let's think about angle B. Since the angles in a triangle add up to 180 degrees, and angle C is 90 degrees, we know that A + B = 90 degrees. This means angles A and B are complementary angles. There's a cool relationship between trigonometric functions of complementary angles:

  • sin A = cos B
  • cos A = sin B
  • tan A = 1 / tan B

So, we can easily find the values for angle B:

  • sin B = cos A = 4/5
  • cos B = sin A = 3/5
  • tan B = 1 / tan A = 4/3

This problem shows how powerful the definition of sine, the Pythagorean theorem, and the relationships between trigonometric functions can be. Even starting with just one trigonometric value, we were able to figure out all the sides and all the trigonometric functions for the angles in the triangle! Remember to always visualize the triangle and use the definitions of the trigonometric ratios. You've got this!

Wrapping Up

So, there you have it, guys! We've worked through three different trigonometry problems, each highlighting a key concept. We used the definitions of sine, cosine, and tangent, the Pythagorean theorem, and the relationships between trigonometric functions to solve these problems. Remember, practice makes perfect, so keep working on these types of problems, and you'll become a trigonometry whiz in no time! Keep up the awesome work, and I'll catch you in the next one!