Question

The slope of a ladder making an angle $ {60^ \circ } $ with floor is ___

A. 1
B. $ - \sqrt 3 $
C. $ - \dfrac{1}{{\sqrt 3 }} $
D. $ \sqrt 3 $

Answer 1

Hint: As we can see in the diagram the vertical change is the opposite side of the angle $ {60^ \circ } $ and horizontal change is the adjacent side to the angle $ {60^ \circ } $ . The ratio of vertical change and horizontal change is the slope which is equal to the ratio of opposite side and adjacent side to the angle $ {60^ \circ } $ . We already know that the ratio of opposite side and adjacent side in a right angled triangle is tangent to the angle. So find tangent to the angle $ {60^ \circ } $ to get the value of slope.

Complete step-by-step answer:
We are given to find the slope of a ladder making an angle $ {60^ \circ } $ with the floor.
Slope is a number which describes both the steepness and direction of a line. It is calculated by dividing the vertical change with horizontal change between any two distinct points on a line.

From the diagram, we can say that slope is $ \dfrac{{\Delta x}}{{\Delta y}} $
But as we can see the ladder with the floor and normal forms a right angled triangle.
In a right triangle, the ratio of opposite side to the adjacent side of an angle gives a tangent function.
Here $ \Delta x $ is the opposite side and $ \Delta y $ is the adjacent side to the angle $ {60^ \circ } $
Therefore, Slope is $ \dfrac{{\Delta x}}{{\Delta y}} = \tan {60^ \circ } $
The value of $ \tan {60^ \circ } $ is $ \sqrt 3 $
Therefore, the slope of the ladder making an angle $ {60^ \circ } $ with floor is $ \sqrt 3 $
So, the correct answer is “Option D”.

Note: Another approach for finding the value of $ \tan {60^ \circ } $
Tangent function is the ratio of sine function to the cosine function.
So to find $ \tan {60^ \circ } $ we need the values of $ \sin {60^ \circ } $ and $ \cos {60^ \circ } $
 $ \sin {60^ \circ } = \dfrac{{\sqrt 3 }}{2},\cos {60^ \circ } = \dfrac{1}{2} $
 $ \tan {60^ \circ } = \dfrac{{\left( {\dfrac{{\sqrt 3 }}{2}} \right)}}{{\left( {\dfrac{1}{2}} \right)}} = \dfrac{{\sqrt 3 }}{1} = \sqrt 3 $