A toy leaves the earth at a point A and rises vertically at uniform speed. After two minutes of vertical rise boy finds the angular elevation of the balloon as 60°.
If the point at which boy is standing is 150 m away from point A, what is the speed of the toy ?

Answer: Option E

Explanation:

Let boy is standing at C position and A is that position from where toy left eartg and B is the position of the toy after 2 minutes.

\begin{aligned}
\text{ Given that CA = 150 m } \\
\text{Angle is }\angle{ 60 ^{\circ} } \\
tan 60 ^{\circ} = \frac{BA}{CA} \\
BA = 150 \sqrt{3}
\end{aligned}
So distance travelled by toy is,
\begin{aligned}
150 \sqrt{3} \\
\text { Total time taken is = 2 min} \\
= \text {2 * 60 = 120 seconds } \\
Speed = \frac{Distance}{Time} \\
= \frac{150\sqrt{3}}{120} = 1.25\sqrt{3}\\
= \text {1.25 * 1.73 = 2.16 mtr/sec}
\end{aligned}

Answer: Option D

Explanation:

Please refer to the image. One of the AB, AD and CD must be given. So data is inadequate.

Answer: Option C

Explanation:

Let AB be the tower and C and D be the positions of the boat.
\begin{aligned}
then \angle{ACB}=45^{\circ}, \angle{ADB}=30^{\circ}, \\
AC = 60&m \\
Let& AB = h \\
=>\frac{AB}{AC} = tan&45^{\circ} = 1 \\
=> AB = AC = 60&m \\

and, \frac{AB}{AD} = tan&30^{\circ} = \frac{1}{\sqrt{3}} \\
=> AD = AB*\sqrt{3} = 60\sqrt{3}&m \\

=> CD = AD-AC = 60\sqrt{3}-60\\
=> CD = 60(\sqrt{3}-1) m \\
\text{We Know Speed = }\frac{Distance}{Time} \\
=> Speed = \left[\frac{60(\sqrt{3}-1)}{5} \right]m/sec
= 12*0.73&m/sec \\

\text{Please note answer we need in Km/Hr} \\
=> Speed = 12*0.73*\frac{18}{5} Km/Hr \\
=> Speed = 31.5 & Km/hr, \\
\text{Which is approx 32 Km/Hr}


\end{aligned}

Answer: Option B

Explanation:

Let AB be the lighthouse and C and D be the positions of the ships.
\begin{aligned}
\text{AB = 100m}, \angle{ACB}=30^{\circ}, \\
\angle{ADB}=45^{\circ}\\
\frac{AB}{AC} = tan&30{\circ} = \frac{1}{\sqrt{3}} \\
=> AC = AB*\sqrt{3} = 100\sqrt{3}m\\

\frac{AB}{AD} = tan&45^{\circ} = 1 \\
=> AB = AD = 100m\\

CD = AC+AD\\
= (100\sqrt{3}+100)&m \\
= 100(\sqrt{3}+1)&m\\
= 100*2.73&m
= 273m

\end{aligned}

Answer: Option D

Explanation:

Let AB be the tower and CD be the electric pole.

\begin{aligned}
\angle{ACB}=60^{\circ}, \angle{EDB}=30^{\circ}, \\
AB = 15&m \\
Let & CD = h, \text{ then}, \\
BE = AB-AE = AB - AE = 15-h\\
\frac{AB}{AC} = tan 60^{\circ} = \sqrt{3} \\
=> AC = \frac{AB}{\sqrt{3}} \\
=> AC = \frac{15}{\sqrt{3}} \\
and, & \frac{BE}{DE} = tan 30^{\circ} = \frac{1}{\sqrt{3}} \\
=> DE = (BE*\sqrt{3}) \\
= \sqrt{3}(15-h) \\

Now, & AC = DE \\
=> \frac{15}{\sqrt{3}} = \sqrt{3}(15-h) \\
=> 3h = 45-15 \\
=> h = \frac{30}{3} = 10 m
\end{aligned}