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exis [7]
3 years ago
6

Help me with number 13 please

Mathematics
1 answer:
Colt1911 [192]3 years ago
6 0
R (-6, 3)
S (-4, 6)
T (1, 4)
U (-4, -1)
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from \: the \: figure \\ AB = 300 \: ft \:  \:  \: angle \: C = 60  \:( theta)\: degress \\ in \: triangle \: ABC \: using \:  tan \: theta\\  \tan(60)  =  \frac{opposite \: side}{adjacent \: side}  \\  \tan(60)  =  \frac{300}{BC}  \\  \sqrt{3}  =  \frac{300}{BC}  \\ BC =  \frac{300}{ \sqrt{3} }  \\ BC = 100 \sqrt{3} ft \\ then \: in \: triangle \: ABD \: using \: tan \: theta \\ angle \: D = 30 \: degrees \: (theta) \\   \tan(30 )  =\frac{opposite \: side}{adjacent \: side} \\  \tan(30)   =  \frac{300}{BD}  \\  \frac{1}{ \sqrt{ 3} }  =  \frac{300}{BD}  \\ BD = 300 \sqrt{3}  \\ CD = BD - BC \\  = 300 \sqrt{3 }  - 100 \sqrt{3}  \\  = (300 - 100) \sqrt{3}  \\  = 200 \sqrt{3 }  \\  = 200 \times 1.732 \\  = 346.4 \: ft

<em>HAVE A NICE DAY</em><em>!</em>

<em>THANKS FOR GIVING ME THE OPPORTUNITY</em><em> </em><em>TO ANSWER YOUR QUESTION</em><em>. </em>

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