The Top of a 15 metre high tower makes an angle of elevation of 60 degree with the bottom of an electric pole and angle of elevation of 30 degree with the top of pole. Find the height of the electric pole.

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 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.
\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 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}