Answer

Verified

70.5k+ views

**Hint:**Please see the below image for reference. Let \[P\] be the position of the man on the top of the tower and \[Q\] be the foot of the tower. Then the points \[A,B,Q\] are collinear and \[PQ \bot AQ\]. Let \[AB = x\], \[BQ = y\], \[PQ = h\]. By the condition, \[\angle PAQ = 30^\circ \] and \[\angle PBQ = 45^\circ \]. The triangles \[PAQ\] and \[PBQ\] are right-angled. First, use the trigonometric ratios to find \[y\] in terms of \[x\] and to calculate the speed of the boat. Then using the relationship between speed, distance, and time, find the required time.

**Formula Used:**

\[\tan \theta = \dfrac{{height}}{{base}}\]

\[speed = \dfrac{{dis\tan ce}}{{time}}\]

**Complete step-by-step answer:**

Given that \[A\] is the initial position of the boat speeding towards a tower and \[B\] is the position of the boat after sailing for \[20\] seconds towards the base of the tower.

Let \[P\] be the position of the man on the top of the tower and \[Q\] be the foot of the tower, which is at the level of water.

Then the points \[A,B,Q\] are collinear and \[PQ \bot AQ\]

Let \[x\] be the distance of the two points \[A\] and \[B\] and \[y\] be the distance of the point \[B\] from the foot of the tower \[Q\].

Also, let \[h\] be the height of the tower.

Then \[AB = x\], \[BQ = y\], \[PQ = h\], \[\angle PAQ = 30^\circ \] and \[\angle PBQ = 45^\circ \]

From triangle \[PAQ\],

\[\dfrac{{PQ}}{{AQ}} = \tan 30^\circ \]

\[ \Rightarrow \dfrac{h}{{x + y}} = \dfrac{1}{{\sqrt 3 }}\]

\[ \Rightarrow x + y = \sqrt 3 h - - - - - \left( i \right)\]

From triangle \[PBQ\],

\[\dfrac{{PQ}}{{BQ}} = \tan 45^\circ \]

\[ \Rightarrow \dfrac{h}{y} = 1\]

\[ \Rightarrow h = y - - - - - \left( {ii} \right)\]

From equations \[\left( i \right)\] and \[\left( {ii} \right)\], we get

\[x + y = \sqrt 3 y\]

\[ \Rightarrow \sqrt 3 y - y = x\]

\[ \Rightarrow \left( {\sqrt 3 - 1} \right)y = x\]

\[ \Rightarrow y = \dfrac{x}{{\sqrt 3 - 1}}\]

After rationalization of denominator, we get,

\[ \Rightarrow y = \dfrac{{x\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 - 1} \right)\left( {\sqrt 3 + 1} \right)}}\]

Identity $(a-b)(a+b)=a^2-b^2$

\[ \Rightarrow y = \dfrac{{x\left( {\sqrt 3 + 1} \right)}}{{{{\left( {\sqrt 3 } \right)}^2} - {{\left( 1 \right)}^2}}}\]

\[ \Rightarrow y = \dfrac{{x\left( {\sqrt 3 + 1} \right)}}{{3 - 1}}\]

\[ \Rightarrow y = \dfrac{{x\left( {\sqrt 3 + 1} \right)}}{2}\]

The boat has reached to the point \[B\] from the point \[A\] in \[20\] seconds and \[AB = x\]

\[\therefore \]Speed of the boat is \[v = \dfrac{x}{{20}}\]

\[\therefore \]The time taken by the boat from \[B\] to reach the base of the tower is \[t = \dfrac{y}{v}\]

Substitute \[y = \dfrac{{x\left( {\sqrt 3 + 1} \right)}}{2}\] and \[v = \dfrac{x}{{20}}\]

\[\therefore t = \dfrac{y}{v} = \dfrac{{\left\{ {\dfrac{{x\left( {\sqrt 3 + 1} \right)}}{2}} \right\}}}{{\left( {\dfrac{x}{{20}}} \right)}} = \dfrac{{x\left( {\sqrt 3 + 1} \right)}}{2} \times \dfrac{{20}}{x} = 10\left( {\sqrt 3 + 1} \right)\]

**Hence option D is correct.**

**Note:**To find the required time, we need to divide the distance traveled by the boat by the uniform speed of the boat. Therefore, first we calculate the speed of the boat and then the distance traveled by the boat has been calculated, which is further utilized to obtain the required time.

Recently Updated Pages

Write an article on the need and importance of sports class 10 english JEE_Main

Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main

Choose the one which best expresses the meaning of class 9 english JEE_Main

What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main

A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main

A man stands at a distance of 250m from a wall He shoots class 9 physics JEE_Main

Other Pages

Which is the correct statement As the scharacter of class 11 chemistry JEE_Main

Electric field due to uniformly charged sphere class 12 physics JEE_Main

The reaction of Zinc with dilute and concentrated nitric class 12 chemistry JEE_Main

The positive temperature coefficient of resistance class 11 physics JEE_Main

A shell explodes and many pieces fly in different directions class 11 physics JEE_Main

The bond order of ClO bond in ClO4 ion and the effective class 11 chemistry JEE_Main