Question

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

Answer 1

the 4 one?

Answer 2

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?

Answer 3

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

Answer 4

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

Answer 5

So the answer is 346.4 ft?

Answer 6

yes indeed