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An aircraft is flying at a height of $3400 \mathrm{~m}$ above the ground. If the angle subtended at a ground observation point by the aircraft positions $10.0 \mathrm{~s}$ apart is $30^{\circ}$, what is the speed of the aircraft?
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In Fig, $O$ is the observation point at the ground. $A$ and $B$ are the positions of air craft for which $\angle A O B=30^{\circ}$. Draw a perpendicular $O C$ on $A B$. Here $O C=3400 \mathrm{~m}$ and $\angle A O C$ $=\angle C O B=15^{\circ}$.
$$
\begin{aligned}
&\text { In } \triangle A O C, A C=O C \tan 15^{\circ}=3400 \times 0.2679 \\
&=910.86 \mathrm{~m} . \\
&A B=A C+C B=A C+A C=2 A C=2 \times 910.86 \mathrm{~m} \\
&\text { Speed of the aircraft, } v=\frac{\text { distance } A B}{\text { time }} \\
&\quad=\frac{2 \times 910.86}{10}=182.17 \mathrm{~ms}^{-1}=182.2 \mathrm{~ms}^{-1}
\end{aligned}
$$
$$
\begin{aligned}
&\text { In } \triangle A O C, A C=O C \tan 15^{\circ}=3400 \times 0.2679 \\
&=910.86 \mathrm{~m} . \\
&A B=A C+C B=A C+A C=2 A C=2 \times 910.86 \mathrm{~m} \\
&\text { Speed of the aircraft, } v=\frac{\text { distance } A B}{\text { time }} \\
&\quad=\frac{2 \times 910.86}{10}=182.17 \mathrm{~ms}^{-1}=182.2 \mathrm{~ms}^{-1}
\end{aligned}
$$
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