Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30 degree and 45 degree respectively. If the lighthouse is 100 m high, the distance between the two ships is:

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 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:

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

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}