Finding Sin(P+R) In A Triangle: A Step-by-Step Guide

Hey guys! Let's dive into a cool math problem today. We're going to figure out how to find the value of sin(P+R) in a triangle, given some cool info about its angles. This problem involves a bit of trigonometry and some triangle properties, but don’t worry, we’ll break it down step by step so it's super easy to follow. So, grab your calculators and let's get started!

Understanding the Problem

Before we jump into the solution, let’s make sure we understand what we’re dealing with. The problem tells us that we have a triangle with angles P, Q, and R. We also know that $\sqrt{2} \tan^2 Q - \tan Q = 0$, and angle Q is an acute angle (meaning it's between 0 and 90 degrees). Our mission is to find the value of $\sin (P+R)$.

Key Concepts

To solve this, we'll need to remember a few key concepts:

1. Angles in a Triangle: The sum of the angles in any triangle is always 180 degrees (or π radians). So, we have P + Q + R = 180°.
2. Trigonometric Identities: We'll use the fact that $\sin (180° - x) = \sin x$. This will be super helpful in simplifying our expression.
3. Solving Trigonometric Equations: We'll need to solve the given equation $\sqrt{2} \tan^2 Q - \tan Q = 0$ to find the value of angle Q.

Breaking Down the Given Information

First, let’s focus on the equation $\sqrt{2} \tan^2 Q - \tan Q = 0$. This looks like a quadratic equation in terms of $\tan Q$. We can factor out $\tan Q$ to make it easier to solve. This is a crucial step, guys, as it simplifies the problem significantly.

$\tan Q (\sqrt{2} \tan Q - 1) = 0$

This gives us two possible solutions:

• $\tan Q = 0$
• $\sqrt{2} \tan Q - 1 = 0$ which simplifies to $\tan Q = \frac{1}{\sqrt{2}}$

Evaluating the Solutions

Now, let's evaluate these solutions. If $\tan Q = 0$, then Q would be 0 degrees or a multiple of 180 degrees. However, since Q is an angle in a triangle, it can't be 0, and it definitely can't be 180 degrees (or more). So, we can discard this solution. Remember, the angles in a triangle must be greater than 0 and less than 180 degrees individually.

That leaves us with $\tan Q = \frac{1}{\sqrt{2}}$. This is a more promising solution. We need to find the angle Q that satisfies this. Think about the values of tan for common angles. Do you guys remember the special right triangles?

Finding Angle Q

We know that $\tan 45° = 1$, so $\tan Q = \frac{1}{\sqrt{2}}$ is close, but not quite 45 degrees. Let's think about other angles. We can use the inverse tangent function (arctan or tan⁻¹) to find the angle Q. This is where your calculator comes in handy!

$Q = \arctan(\frac{1}{\sqrt{2}})$

If you calculate this, you’ll find that Q is approximately 35.26 degrees. However, to make our calculations simpler and more precise, let’s think about this in terms of special right triangles. We have a tangent value that's not immediately recognizable, but the key here is to recognize that we’re dealing with an angle whose tangent is $\frac{1}{\sqrt{2}}$.

Let's try to avoid using approximations for now. Instead, let's keep $\tan Q = \frac{1}{\sqrt{2}}$ as it is and see how we can use it. We know that Q is an acute angle, so it lies between 0 and 90 degrees. This is super important because it helps us narrow down the possible solutions.

Finding sin(P+R)

Now, let's get back to what we need to find: $\sin (P+R)$. We know that P + Q + R = 180°, so we can write P + R as 180° - Q. This is a neat trick that allows us to use our trig identities.

P + R = 180° - Q

Now we can substitute this into our expression:

$\sin (P+R) = \sin (180° - Q)$

Using the trigonometric identity $\sin (180° - x) = \sin x$, we get:

$\sin (P+R) = \sin Q$

So, our problem now boils down to finding $\sin Q$. We already know $\tan Q = \frac{1}{\sqrt{2}}$. How can we find $\sin Q$ from this? This is where our knowledge of trigonometric relationships comes into play.

Using Trigonometric Relationships

We can use the Pythagorean identity to relate $\tan Q$ to $\sin Q$ and $\cos Q$. Remember the identity?

$\tan Q = \frac{\sin Q}{\cos Q}$

We also know the fundamental Pythagorean trigonometric identity:

$\sin^2 Q + \cos^2 Q = 1$

We can rewrite the tangent equation as:

$\sin Q = \tan Q \cdot \cos Q = \frac{1}{\sqrt{2}} \cos Q$

Now, let's substitute this into the Pythagorean identity:

$(\frac{1}{\sqrt{2}} \cos Q)^2 + \cos^2 Q = 1$

$\frac{1}{2} \cos^2 Q + \cos^2 Q = 1$

Combining the terms, we get:

$\frac{3}{2} \cos^2 Q = 1$

Now, we can solve for $\cos^2 Q$:

$\cos^2 Q = \frac{2}{3}$

Taking the square root, we get:

$\cos Q = \pm \sqrt{\frac{2}{3}}$

Since Q is an acute angle, $\cos Q$ must be positive. So,

$\cos Q = \sqrt{\frac{2}{3}}$

Now we can find $\sin Q$ using the equation $\sin Q = \frac{1}{\sqrt{2}} \cos Q$:

$\sin Q = \frac{1}{\sqrt{2}} \cdot \sqrt{\frac{2}{3}}$

$\sin Q = \sqrt{\frac{1}{2}} \cdot \sqrt{\frac{2}{3}}$

$\sin Q = \sqrt{\frac{1}{3}}$

$\sin Q = \frac{1}{\sqrt{3}}$

Rationalizing the denominator, we get:

$\sin Q = \frac{\sqrt{3}}{3}$

Final Answer

Remember, we found that $\sin (P+R) = \sin Q$. So,

$\sin (P+R) = \frac{\sqrt{3}}{3}$

And that's our final answer! Guys, we did it! We found the value of $\sin (P+R)$ by using the given information, trigonometric identities, and a bit of algebraic manipulation. Pat yourselves on the back!

Conclusion

So, there you have it! We've successfully navigated through a tricky trigonometry problem by breaking it down into manageable steps. Remember, the key is to understand the underlying concepts, use the given information wisely, and apply the right identities. Practice makes perfect, so keep tackling those problems, and you'll become a math whiz in no time! Keep up the great work, everyone! You got this!