If (a cos theta_i, a sin theta_i) for i = 1, | vedclass

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(A) The vertices of the triangle are $A(a \cos 	heta_1, a \sin 	heta_1)$,$B(a \cos 	heta_2, a \sin 	heta_2)$,and $C(a \cos 	heta_3, a \sin 	heta

Vedclass · Accepted Answer

$\cos 	heta_1 + \cos 	heta_2 + \cos 	heta_3 = 0$

Vedclass · Answer

(A) The vertices of the triangle are $A(a \cos 	heta_1, a \sin 	heta_1)$,$B(a \cos 	heta_2, a \sin 	heta_2)$,and $C(a \cos 	heta_3, a \sin 	heta_3)$.Since the triangle is equilateral and inscribed in the circle $x^2 + y^2 = a^2$,the centroid of the triangle coincides with the center of the circle,which is $(0, 0)$.The centroid $(G)$ of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3})$.Setting the centroid to $(0, 0)$,we get:$\frac{a \cos 	heta_1 + a \cos 	heta_2 + a \cos 	heta_3}{3} = 0 \implies \cos 	heta_1 + \cos 	heta_2 + \cos 	heta_3 = 0$$\frac{a \sin 	heta_1 + a \sin 	heta_2 + a \sin 	heta_3}{3} = 0 \implies \sin 	heta_1 + \sin 	heta_2 + \sin 	heta_3 = 0$Thus,the correct option is $A$.

If $(a \cos \theta_i, a \sin \theta_i)$ for $i = 1, 2, 3$ represent the vertices of an equilateral triangle inscribed in a circle $x^2 + y^2 = a^2$,then:

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