Nice Geometry Problem #6

Geometry Level 2

\frac{8}{3}

\frac{16}{3}

\frac{64}{3}

\frac{27}{2}

\frac{9}{2}

\frac{32}{3}

\frac{3}{2}

\frac{81}{2}

2 solutions

Chew-Seong Cheong
Apr 17, 2018

We note that $\sin A = \dfrac {BD}{AB} = \dfrac {BD}{24} = \dfrac 23$ $\implies BD = \dfrac 23 \times 24 = 16$ . Also $\sin C = \dfrac {BD}{BC} = \dfrac {16}{BC} = \dfrac 34$ $\implies BC = 16 \times \dfrac 43 = \boxed{\dfrac {64}3}$ .

César Castro
Mar 30, 2018

Nice Geometry Problem #6

2 solutions

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