Question

A ladder is resting with the wall at an angle of 30 degrees. A man is ascending the ladder at the rate of 3ft/sec. Find his rate of approaching the wall.
A. 3ft/sec
B. \[\dfrac{3}{2}\] ft/sec
C. \[\dfrac{2}{4}\]ft/sec
D. \[\dfrac{3}{{\sqrt 2 }}\] ft/sec

Answer 1

Hint: Man is ascending the ladder at the rate of 3ft/s means that the slant speed is 3ft/s. And how fast he is approaching means horizontal speed. So, we had to find the horizontal speed if the slant speed was given.

Formula used
The formula of horizontal speed is=
Slant speed\[ \times \cos \theta \] , where \[\theta \] is the angle made by the base with slant.
The formula of vertical speed is=
Slant speed\[ \times \sin \theta \],where \[\theta \] is the angle made by the perpendicular line with slant.

Complete step by step solution
Let us first draw the diagram of the given problem,

As we can see from the above figure that man is ascending the ladder at a speed of 3ft/s or we can say that the speed along line BA is 3ft/s.
Now we are asked to find the speed of man along line BC.
So, as we know that walls and ground are always perpendicular to each other. So, the angle ACB must be equal to \[{90^ \circ }\].
As we know that the sum of all angles of a triangle is equal to \[{180^ \circ }\] and the angle CAB (i.e. angle made by ladder with the wall) is given as \[{30^ \circ }\].
So, the angle ABC =
\[{180^ \circ } - {90^ \circ } - {30^ \circ }\]
\[ = {60^ \circ }\] .

Now as we know that if the slant speed is x then the horizontal speed will be \[x\cos \theta \].
\[\theta \] is the angle made by the base with slant and vertical speed will be \[x\sin \theta \], where \[\theta \] is the angle made by the perpendicular line with a slant.
Hence, the horizontal speed will be equal to \[3\cos {60^ \circ }\] .
Now as we know that \[\cos {60^ \circ } = \dfrac{1}{2}\]
So, the horizontal speed will be = \[\dfrac{3}{2}\] ft/sec
So, the man is approaching the wall at the speed of \[\dfrac{3}{2}\] ft/sec.
The correct option is B.

Note Whenever we come up with this type of problem then first, we had to draw the diagram because it makes the problem easier to understand and after that, if we know the slant speed (like the speed at which man is ascending wall) and asked to find horizontal speed (like how fast man approaching wall) then we can direct use formula horizontal speed = slant speed* \[\cos \theta \], where \[\theta \] will be the angle between slant speed and horizontal speed. This will be the easiest and most efficient way to find the solution to the problem.