Explore Westonci.ca, the premier Q&A site that helps you find precise answers to your questions, no matter the topic. Explore our Q&A platform to find reliable answers from a wide range of experts in different fields. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
Alright, let's break down the solution step-by-step to find the angle opposite the leg that is 6 inches long in this right triangle.
1. Identify the given information:
- One leg (let's call it [tex]\( a \)[/tex]) of the right triangle measures 6 inches.
- The other leg (let's call it [tex]\( b \)[/tex]) of the right triangle measures [tex]\( 6\sqrt{3} \)[/tex] inches.
2. Understand the problem:
- We need to find the measure of the angle opposite the leg that is 6 inches long.
3. Use the basic trigonometric ratio:
- In a right triangle, the tangent of an angle is the ratio of the length of the opposite leg to the length of the adjacent leg.
- So, for the angle [tex]\( \theta \)[/tex] opposite the leg measuring 6 inches:
[tex]\[ \tan(\theta) = \frac{\text{opposite leg}}{\text{adjacent leg}} = \frac{a}{b} = \frac{6}{6\sqrt{3}} \][/tex]
4. Simplify the fraction:
- To simplify the fraction [tex]\( \frac{6}{6\sqrt{3}} \)[/tex], divide the numerator and the denominator by 6:
[tex]\[ \frac{6}{6\sqrt{3}} = \frac{1}{\sqrt{3}} \][/tex]
- We can further simplify [tex]\( \frac{1}{\sqrt{3}} \)[/tex] by rationalizing the denominator:
[tex]\[ \frac{1}{\sqrt{3}} = \frac{1 \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}} = \frac{\sqrt{3}}{3} \][/tex]
5. Determine the angle:
- We now know that:
[tex]\[ \tan(\theta) = \frac{\sqrt{3}}{3} \][/tex]
- In trigonometry, the angle [tex]\( \theta \)[/tex] whose tangent is [tex]\( \frac{\sqrt{3}}{3} \)[/tex] is [tex]\( 30^\circ \)[/tex].
6. Confirm the angle:
- The tangent function is well-defined and, for common angles, we know that:
[tex]\[ \tan(30^\circ) = \frac{\sqrt{3}}{3} \][/tex]
- Therefore, the angle opposite the leg measuring 6 inches is indeed [tex]\( 30^\circ \)[/tex].
So, the measure of the angle opposite the leg that is 6 inches long is:
[tex]\[ \boxed{30^\circ} \][/tex]
1. Identify the given information:
- One leg (let's call it [tex]\( a \)[/tex]) of the right triangle measures 6 inches.
- The other leg (let's call it [tex]\( b \)[/tex]) of the right triangle measures [tex]\( 6\sqrt{3} \)[/tex] inches.
2. Understand the problem:
- We need to find the measure of the angle opposite the leg that is 6 inches long.
3. Use the basic trigonometric ratio:
- In a right triangle, the tangent of an angle is the ratio of the length of the opposite leg to the length of the adjacent leg.
- So, for the angle [tex]\( \theta \)[/tex] opposite the leg measuring 6 inches:
[tex]\[ \tan(\theta) = \frac{\text{opposite leg}}{\text{adjacent leg}} = \frac{a}{b} = \frac{6}{6\sqrt{3}} \][/tex]
4. Simplify the fraction:
- To simplify the fraction [tex]\( \frac{6}{6\sqrt{3}} \)[/tex], divide the numerator and the denominator by 6:
[tex]\[ \frac{6}{6\sqrt{3}} = \frac{1}{\sqrt{3}} \][/tex]
- We can further simplify [tex]\( \frac{1}{\sqrt{3}} \)[/tex] by rationalizing the denominator:
[tex]\[ \frac{1}{\sqrt{3}} = \frac{1 \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}} = \frac{\sqrt{3}}{3} \][/tex]
5. Determine the angle:
- We now know that:
[tex]\[ \tan(\theta) = \frac{\sqrt{3}}{3} \][/tex]
- In trigonometry, the angle [tex]\( \theta \)[/tex] whose tangent is [tex]\( \frac{\sqrt{3}}{3} \)[/tex] is [tex]\( 30^\circ \)[/tex].
6. Confirm the angle:
- The tangent function is well-defined and, for common angles, we know that:
[tex]\[ \tan(30^\circ) = \frac{\sqrt{3}}{3} \][/tex]
- Therefore, the angle opposite the leg measuring 6 inches is indeed [tex]\( 30^\circ \)[/tex].
So, the measure of the angle opposite the leg that is 6 inches long is:
[tex]\[ \boxed{30^\circ} \][/tex]
We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.