A man standing at a point C is watching the top of a tower, which makes an angle of elevation of 30 degree. The man walks some distance towards the tower to watch its top and the angle of elevation become 60 degree. What is the distance between the base of the tower and point C.

Answer: Option C

Explanation:

Let AB be the tower.
\begin{aligned}
then \angle{ACB}= 30^{\circ} \\
AB = 100&meter \\
\frac{AB}{AC} = tan&30^{\circ} \\
=>\frac{100}{AC} = \frac{1}{\sqrt{3}} \\
=> AC = \sqrt{3}*100 \\
=> AC = 1.73*100 \\
=> AC = 173 meter
\end{aligned}

Please always remember the value of square root 3 is 1.73, and value of square root 2 is 1.41.
This will be very helpful while solving height and distance questions and saving your time.

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 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}