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

the 4 one?

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?

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

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

So the answer is 346.4 ft?

yes indeed

Join our real-time social learning platform and learn together with your friends!