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Answer
(i) (b): From Fig. we find that $A B=200 m$ and $\theta=30^{\circ}$. In right triangle $A B C$, we obtain
$
\sin \theta=\frac{A B}{A C} \Rightarrow \sin 30^{\circ}=\frac{200}{A C} \Rightarrow \frac{1}{2}=\frac{200}{A C} \Rightarrow A C=400 m
$
(ii) (c): We have, $A B=200 m$ and $\theta=30^{\circ}$. In right triangle $A B C$, we obtain
$
\tan 30^{\circ}=\frac{A B}{B C} \Rightarrow \frac{1}{\sqrt{3}}=\frac{200}{B C} \Rightarrow B C=200 \sqrt{3} m
$
(iii) (d): If $B C=15 m$ and $\theta=30^{\circ}$, then
$
\tan \theta=\frac{A B}{B C} \Rightarrow \tan 30^{\circ}=\frac{A B}{15} \Rightarrow \frac{1}{\sqrt{3}}=\frac{A B}{15} \Rightarrow A B=5 \sqrt{3} m
$
(iv) (c): In $\triangle A B C$, we obtain
$
\tan \theta=\frac{A B}{B C} \Rightarrow \tan \theta=1 \Rightarrow \tan \theta=\tan 45^{\circ} \Rightarrow \theta=45^{\circ}
$
$
\sin \theta=\frac{A B}{A C} \Rightarrow \sin 30^{\circ}=\frac{200}{A C} \Rightarrow \frac{1}{2}=\frac{200}{A C} \Rightarrow A C=400 m
$
(ii) (c): We have, $A B=200 m$ and $\theta=30^{\circ}$. In right triangle $A B C$, we obtain
$
\tan 30^{\circ}=\frac{A B}{B C} \Rightarrow \frac{1}{\sqrt{3}}=\frac{200}{B C} \Rightarrow B C=200 \sqrt{3} m
$
(iii) (d): If $B C=15 m$ and $\theta=30^{\circ}$, then
$
\tan \theta=\frac{A B}{B C} \Rightarrow \tan 30^{\circ}=\frac{A B}{15} \Rightarrow \frac{1}{\sqrt{3}}=\frac{A B}{15} \Rightarrow A B=5 \sqrt{3} m
$
(iv) (c): In $\triangle A B C$, we obtain
$
\tan \theta=\frac{A B}{B C} \Rightarrow \tan \theta=1 \Rightarrow \tan \theta=\tan 45^{\circ} \Rightarrow \theta=45^{\circ}
$
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