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 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 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 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}

Answer: Option A

Explanation:

Let AB be the tree and AC be its shadow, So as per question, shadow of a tree is \begin{aligned}\sqrt{3}\end{aligned} times the height of tree. Let h be the height of the tree, then
\begin{aligned}
\frac{AB}{AC} = tan\theta \\
=> \frac{h}{\sqrt{3}h} = tan\theta \\
=> tan\theta = \frac{1}{\sqrt{3}} \\
=> \theta = 30^{\circ}
\end{aligned}

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}