Question

the centroid of a triangle formed by the points (0,0), (cos a, sin a), and (sin a, -cos a) lies on the line y 2x, then a-? [imagine a is teta

Answer 1

Answer on Question #83988 – Math – Geometry Question

The centroid of a triangle formed by the points $(0,0)$ , (cos a, sin a), and (sin a, -cos a) lies on the line $y = 2x$ , then a-?

Solution

The centroid is the intersection point of the medians. Line $y = 2x$ goes through the point $(0,0)$ which is one of the vertices of the triangle. And also this line goes through the centroid by condition. This means that line $y = 2x$ is one of the medians of triangle.

Find the middle of the segment (cos a, sin a) (sin a, -cos a):

$$\left(\left(\cos a + \sin a\right) / 2, \left(\sin a - \cos a\right) / 2\right)$$

This point lies on the line $y = 2x$ which means:

$$\begin{array}{l} (\sin a - \cos a) / 2 = 2 * (\cos a + \sin a) / 2 \\ \sin a - \cos a = 2 \cos a + 2 \sin a \\ 3 \cos a + \sin a = 0 \\ 3 \cos a + \sqrt {(1 - \cos^ {2} a)} = 0 \\ (1 - \cos^ {2} a) = 9 \cos^ {2} a \\ \cos^ {2} a = 1 / 1 0 \\ \cos a = \pm 1 / \sqrt {1 0} \\ \end{array}$$

We got two solutions:

$$\cos a = 1 / \sqrt {1 0}, \sin a = - 3 / \sqrt {1 0}$$ $$\cos a = - 1 / \sqrt {1 0}, \sin a = 3 / \sqrt {1 0}$$

Draw solutions on the chart and find the angle $a$ .

$$\cos a = 1 / \sqrt {1 0}, \sin a = - 3 / \sqrt {1 0}$$

\cos a = -1 / \sqrt{10}

sin

a = 3 / \sqrt{10}

Point (cos a, sin a) (red) is in the second quarter:

$$a = \cos^ {- 1} (- 1 / \sqrt {1 0}) \approx 1. 8 9 2 5 \mathrm {r a d}$$

Answer: $2\pi - \cos^{-1}(1 / \sqrt{10}) \approx 5.0341$ rad, $\cos^{-1}(-1 / \sqrt{10}) \approx 1.8925$ rad.

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