Answer :
To find the sine value of the given point \((-6, 5)\) on a coordinate plane, we need to understand the basic properties of right triangles and trigonometry concepts.
First, let's identify the key elements:
- The point \((-6, 5)\) represents the coordinates \((x, y)\).
- \(x\) is the horizontal distance from the origin, and \(y\) is the vertical distance from the origin.
The problem requires us to find the sine of the angle \(\theta\) formed by the radius \(r\) (the hypotenuse) connecting the origin (0, 0) to the point \((-6, 5)\).
### Steps to Solve:
1. Determine the hypotenuse \(r\):
The hypotenuse \(r\) of a right triangle can be found using the Pythagorean theorem, which states \(r = \sqrt{x^2 + y^2}\).
2. Calculate \(r\):
[tex]\[ r = \sqrt{(-6)^2 + 5^2} \][/tex]
Breaking it down:
[tex]\[ (-6)^2 = 36 \][/tex]
[tex]\[ 5^2 = 25 \][/tex]
[tex]\[ r = \sqrt{36 + 25} = \sqrt{61} \][/tex]
3. Determine the sine value:
The sine of an angle in a right triangle is defined as the ratio of the length of the opposite side to the hypotenuse. Here, the opposite side is \(y = 5\) and the hypotenuse is \(r\).
4. Calculate the sine value:
[tex]\[ \sin(\theta) = \frac{y}{r} = \frac{5}{\sqrt{61}} \][/tex]
5. Rationalize the denominator (if necessary):
To rationalize \(\frac{5}{\sqrt{61}}\), multiply the numerator and the denominator by \(\sqrt{61}\):
[tex]\[ \sin(\theta) = \frac{5 \sqrt{61}}{61} \][/tex]
### Conclusion:
The sine value for the point \((-6, 5)\) is:
[tex]\[ \sin(\theta) = \frac{5 \sqrt{61}}{61} \][/tex]
Thus, the correct answer from the given options is:
[tex]\[ \boxed{\frac{5 \sqrt{61}}{61}} \][/tex]
First, let's identify the key elements:
- The point \((-6, 5)\) represents the coordinates \((x, y)\).
- \(x\) is the horizontal distance from the origin, and \(y\) is the vertical distance from the origin.
The problem requires us to find the sine of the angle \(\theta\) formed by the radius \(r\) (the hypotenuse) connecting the origin (0, 0) to the point \((-6, 5)\).
### Steps to Solve:
1. Determine the hypotenuse \(r\):
The hypotenuse \(r\) of a right triangle can be found using the Pythagorean theorem, which states \(r = \sqrt{x^2 + y^2}\).
2. Calculate \(r\):
[tex]\[ r = \sqrt{(-6)^2 + 5^2} \][/tex]
Breaking it down:
[tex]\[ (-6)^2 = 36 \][/tex]
[tex]\[ 5^2 = 25 \][/tex]
[tex]\[ r = \sqrt{36 + 25} = \sqrt{61} \][/tex]
3. Determine the sine value:
The sine of an angle in a right triangle is defined as the ratio of the length of the opposite side to the hypotenuse. Here, the opposite side is \(y = 5\) and the hypotenuse is \(r\).
4. Calculate the sine value:
[tex]\[ \sin(\theta) = \frac{y}{r} = \frac{5}{\sqrt{61}} \][/tex]
5. Rationalize the denominator (if necessary):
To rationalize \(\frac{5}{\sqrt{61}}\), multiply the numerator and the denominator by \(\sqrt{61}\):
[tex]\[ \sin(\theta) = \frac{5 \sqrt{61}}{61} \][/tex]
### Conclusion:
The sine value for the point \((-6, 5)\) is:
[tex]\[ \sin(\theta) = \frac{5 \sqrt{61}}{61} \][/tex]
Thus, the correct answer from the given options is:
[tex]\[ \boxed{\frac{5 \sqrt{61}}{61}} \][/tex]
