A man on the top of a vertical observation tower observes a car moving at a
uniform speed coming directly towards it. If it takes 12 minutes for the angle of
depression to change from \[{30^0}\] to ${45^0}$. How soon after this will the car reach the
observation tower?
\[
(A){\text{ 14min 3 sec}} \\
(B){\text{ 15min 49sec}} \\
(C){\text{ 16min 23sec}} \\
(D){\text{ 18min 5sec}} \\
\]
Last updated date: 19th Mar 2023
•
Total views: 306.6k
•
Views today: 4.85k
Answer
306.6k+ views
Hint: Draw figure and then use trigonometry identity $\tan \theta = \dfrac{{Perpendicular}}{{Base}}$.
Above figure is drawn with respect to the given conditions in question.
As we can see from the above figure that,
Man is on the top of a vertical tower.
And to change angle of depression from \[{30^0}\] to ${45^0}$ i.e. \[\angle {\text{ADB}}\] to \[\angle {\text{ACB}}\].
It takes 12 minutes,
And it is obvious that when the car will reach the observation tower,
then the angle of depression will be ${90^0}$.
Let the height of the tower be $y$ units.
As we are given that the time taken to travel DC (see in figure) is 12 minutes.
Let the time taken to travel CB will be $x$ minutes.
Here we are known with perpendicular and base of \[\Delta {\text{ABC}}\] and \[\Delta {\text{ABD}}\].
So, we will only use that trigonometric functions, that include perpendicular and base
So, as we know that, $\tan \theta = \dfrac{{Perpendicular}}{{Base}}$.
So, as we can see from the above figure, $\tan {45^0} = \dfrac{{AB}}{{CB}} = \dfrac{y}{x}$.
So, $x = y$ ……………………………………….(1)
And, $\tan 30^\circ = \dfrac{{AB}}{{DB}} = \dfrac{{AB}}{{DC + CB}} = \dfrac{y}{{12 + x}}$.
Now, putting the value of $\tan {30^0}$ and $y$ from equation 1. We get,
$\dfrac{1}{{\sqrt 3 }} = \dfrac{x}{{12 + x}} \Rightarrow \left( {\sqrt 3 - 1} \right)x = 12 \Rightarrow x = \dfrac{{12}}{{\left( {\sqrt 3 - 1} \right)}} \approx 16.38$minutes
Now, as we have defined above that time taken to travel CB is x minutes.
So, time taken to reach the observation tower will be x minutes.
So, according to the options given in the question
the most appropriate answer will be 16min 23 sec.
Hence, the correct Option will be C.
Note: Whenever we come up with these types of problems first, we should draw a figure according to the given conditions in question. And then we will assume time taken to reach the tower as x and then after using trigonometric functions like \[{\text{tan}}\theta \], we can get the value of x using angle of depression and time taken to change angle of depression. This will be the easiest and efficient way to reach the required solution of the problem.

Above figure is drawn with respect to the given conditions in question.
As we can see from the above figure that,
Man is on the top of a vertical tower.
And to change angle of depression from \[{30^0}\] to ${45^0}$ i.e. \[\angle {\text{ADB}}\] to \[\angle {\text{ACB}}\].
It takes 12 minutes,
And it is obvious that when the car will reach the observation tower,
then the angle of depression will be ${90^0}$.
Let the height of the tower be $y$ units.
As we are given that the time taken to travel DC (see in figure) is 12 minutes.
Let the time taken to travel CB will be $x$ minutes.
Here we are known with perpendicular and base of \[\Delta {\text{ABC}}\] and \[\Delta {\text{ABD}}\].
So, we will only use that trigonometric functions, that include perpendicular and base
So, as we know that, $\tan \theta = \dfrac{{Perpendicular}}{{Base}}$.
So, as we can see from the above figure, $\tan {45^0} = \dfrac{{AB}}{{CB}} = \dfrac{y}{x}$.
So, $x = y$ ……………………………………….(1)
And, $\tan 30^\circ = \dfrac{{AB}}{{DB}} = \dfrac{{AB}}{{DC + CB}} = \dfrac{y}{{12 + x}}$.
Now, putting the value of $\tan {30^0}$ and $y$ from equation 1. We get,
$\dfrac{1}{{\sqrt 3 }} = \dfrac{x}{{12 + x}} \Rightarrow \left( {\sqrt 3 - 1} \right)x = 12 \Rightarrow x = \dfrac{{12}}{{\left( {\sqrt 3 - 1} \right)}} \approx 16.38$minutes
Now, as we have defined above that time taken to travel CB is x minutes.
So, time taken to reach the observation tower will be x minutes.
So, according to the options given in the question
the most appropriate answer will be 16min 23 sec.
Hence, the correct Option will be C.
Note: Whenever we come up with these types of problems first, we should draw a figure according to the given conditions in question. And then we will assume time taken to reach the tower as x and then after using trigonometric functions like \[{\text{tan}}\theta \], we can get the value of x using angle of depression and time taken to change angle of depression. This will be the easiest and efficient way to reach the required solution of the problem.
Recently Updated Pages
Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts
Name the Largest and the Smallest Cell in the Human Body ?

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

A ball impinges directly on a similar ball at rest class 11 physics CBSE

Lysosomes are known as suicidal bags of cell why class 11 biology CBSE

Two balls are dropped from different heights at different class 11 physics CBSE

A 30 solution of H2O2 is marketed as 100 volume hydrogen class 11 chemistry JEE_Main

A sample of an ideal gas is expanded from 1dm3 to 3dm3 class 11 chemistry CBSE
