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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?
\[
(A){\text{ 14min 3 sec}} \\
(B){\text{ 15min 49sec}} \\
(C){\text{ 16min 23sec}} \\
(D){\text{ 18min 5sec}} \\
\]
Answer
125.1k+ views
Hint: Draw figure and then use trigonometry identity $\tan \theta = \dfrac{{Perpendicular}}{{Base}}$.
![](https://www.vedantu.com/question-sets/84527ce7-956a-4cf5-86a2-29f9d65f73326947470776291831559.png)
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.
![](https://www.vedantu.com/question-sets/84527ce7-956a-4cf5-86a2-29f9d65f73326947470776291831559.png)
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.
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