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Mathematics kekeman:

More math: https://snipboard.io/qT0jwR.jpg ityourgirlcrystal:

the 4 one? GoodFellow:

You cannot find CD directly, but you probably can find BC & then if u can find BD then CD = BD - BC. Guess isn't it about finding the value of angle A? Florisalreadytaken:

we can see that: $opposite=300 \ ft$ $\theta=60^\circ$ SOH CAH TOA we have to use TOA as we are given the opposite, theta, and just have to find the adjacent: $\tan \theta=\frac{opposite}{adjacent}$ $\tan 60^\circ = \frac{300}{adjacent}$ $adjacent = \frac{300}{01.732}$ $adjacent(BC) \approx 173.2 \ ft$ now lets do the 2nd triangle (the big one) we can see that: $opposite=300 \ ft$ $\theta=30^\circ$ SOH CAH TOA we have to use TOH again $\tan \theta=\frac{opposite}{adjacent}$ $\tan 30^\circ = \frac{300}{adjacent}$ $adjacent = \frac{300}{0.577}$ $adjacent (BD)\approx 519.9$ now to find the the difference of the two what is 519.9-173.2? its an approx answer -- so its not exactly as the option -- but its VERY close to it surjithayer:

$\tan 60=\frac{ 300 }{ BC }$ $BC=\frac{ 300 }{ \tan 60 }$ $=\frac{ 300 }{ \sqrt{3} }\times \frac{ \sqrt{3} }{\sqrt{3} }$ $=100\sqrt{3}~...(1)$ Again $\tan 30=\frac{ 300 }{ BD }$ $BD=\frac{ 300 }{ \tan 30 }$ $=\frac{ 300 }{ \frac{ 1 }{ \sqrt{3} } }=300 \sqrt{3}$ $BC+CD=300\sqrt{3} ...(2)$ $100\sqrt{3}+CD=300\sqrt{3}$ $CD=300\sqrt{3}-100\sqrt{3}=200\sqrt{3}\approx 346.4~ft$ kekeman:

So the answer is 346.4 ft? Florisalreadytaken:

yes indeed

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