1. In triangle ABC , AB=9.2 ft, BC=11.9 ft, and m

1. and m B=27dg. What is the area of ABC? Round to the nearest hundredth

1 is not a question......

2. Call the point Marco stands on "M", Garret stands on "G" and Dino stands on "D".

Apply the law of sines:

17 sin angle MDG = 15 sin angle MGD

because $\angle MDG +\angle MGD = (180 - 81)^{\circ}=99^{\circ}$

$17\sin(99^{\circ}-\angle MGD)=15\sin \angle MGD \\17(\sin99^{\circ}\cos \angle MGD-\cos 99^{\circ}\sin\angle MGD)=15\sin \angle MGD\\ 17\sin 99^{\circ} \cot \angle MGD - 17\cos 99^{\circ} = 15\\ \cot \angle MGD = \dfrac{15 + 17\cos 99^{\circ}}{17\sin 99^{\circ}}\\ \angle MGD \approx 53.69^{\circ}\\ \therefore \angle MDG = (99 - 53.69)^{\circ}=45.31^{\circ}$

DG = 17cos(45.31) + 15cos(53.69) = 21 yd

Area = 1/2 * 9.2 * 11.9 * sin(27 dg) = 24.85 ft

Or use the law of cosines directly:  a² = b² + c² -2b.c.cos(A)

GD² = 17² + 15² -2.17.15.cos(81°) = 434.218 yd²

GD = 20.8 yd or GD = 21 yd to nearest yard.

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