#### Answer

$18.0$ m$^{2}$

#### Work Step by Step

$A=\frac{1}{2}(4)(10)\sin64^{\circ}$
$A=20\sin64^{\circ}$
$A\approx18.0$ m$^{2}$

Published by
Brooks Cole

ISBN 10:
1439047901

ISBN 13:
978-1-43904-790-3

$18.0$ m$^{2}$

$A=\frac{1}{2}(4)(10)\sin64^{\circ}$
$A=20\sin64^{\circ}$
$A\approx18.0$ m$^{2}$

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