Aero - Fly - 2

I pictured this as a right triangle created by the observation point, and the start and finish points of the aircraft.

Let $x$ be the distance the plane travels in $10s$ . Then using the sine rule $\frac { 3400 }{ \sin { 60 } } =\frac { x }{ \sin { 30 } } \\ \Rightarrow x=\frac { 3400\sin { 30 } }{ \sin { 60 } } \\ \Rightarrow x= \frac { 3400\sqrt { 3 } }{ 3 }$ So in the speed of the aircraft (assuming it travels at a constant speed) is $\frac { x }{ t } =\frac { \frac { 3400\sqrt { 3 } }{ 3 } }{ 10 } = \frac { 340\sqrt { 3 } }{ 3 } =196.2990915\dots \simeq \boxed { 196.3{ ms }^{ -1 } } .$

The observer is $3400\space m$ below the aircraft at time $t=0\space s$ , then $t=10\space s$ later, the aircraft is $30^\circ$ from the vertical above the observer.

Therefore, the distance traveled by the aircraft $d = 3400 \tan{30^\circ} = \dfrac {3400}{\sqrt{3}}$ and the speed of the aircraft $v = \dfrac{d}{10} = \dfrac {340}{\sqrt{3}} = \boxed{196.3}\space s$ .