Let A(6,8),B(10 cos alpha, -10 sin alpha) and | vedclass

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(B) The vertices $A(6, 8)$,$B(10 \cos \alpha, -10 \sin \alpha)$,and $C(-10 \sin \alpha, 10 \cos \alpha)$ all lie on the circle $x^2 + y^2 = 100$. Thus

Vedclass · Accepted Answer

$145$

Vedclass · Answer

(B) The vertices $A(6, 8)$,$B(10 \cos \alpha, -10 \sin \alpha)$,and $C(-10 \sin \alpha, 10 \cos \alpha)$ all lie on the circle $x^2 + y^2 = 100$. Thus,the circumcenter $O$ is $(0, 0)$.In any triangle,the orthocenter $L$,centroid $G$,and circumcenter $O$ are collinear,and $G$ divides $OL$ in the ratio $2:1$. Using the section formula,$G(h, k) = \left(\frac{1 \cdot a + 2 \cdot 0}{1+2}, \frac{1 \cdot 9 + 2 \cdot 0}{1+2}\right) = \left(\frac{a}{3}, 3\right)$.Thus,$h = \frac{a}{3} \Rightarrow a = 3h$ and $k = 3$.Also,the centroid $G(h, k)$ is given by $\left(\frac{6 + 10 \cos \alpha - 10 \sin \alpha}{3}, \frac{8 - 10 \sin \alpha + 10 \cos \alpha}{3}\right) = (h, k)$.Since $k = 3$,we have $\frac{8 + 10(\cos \alpha - \sin \alpha)}{3} = 3 \Rightarrow 10(\cos \alpha - \sin \alpha) = 1$.Squaring both sides,$100(\cos^2 \alpha + \sin^2 \alpha - 2 \sin \alpha \cos \alpha) = 1$ $\Rightarrow 100(1 - \sin 2\alpha) = 1$ $\Rightarrow 100 \sin 2\alpha = 99$.From $h = \frac{6 + 10(\cos \alpha - \sin \alpha)}{3}$,we get $h = \frac{6 + 1}{3} = \frac{7}{3}$.Now,calculate $5a - 3h + 6k + 100 \sin 2\alpha = 5(3h) - 3h + 6(3) + 99 = 12h + 18 + 99 = 12(\frac{7}{3}) + 117 = 28 + 117 = 145$.

Let $A(6,8)$,$B(10 \cos \alpha, -10 \sin \alpha)$ and $C(-10 \sin \alpha, 10 \cos \alpha)$ be the vertices of a triangle. If $L(a, 9)$ and $G(h, k)$ are its orthocenter and centroid respectively,then $(5a - 3h + 6k + 100 \sin 2\alpha)$ is equal to . . . . . .

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