Answer:
[tex]b.\ cos 70 =\dfrac{4}x[/tex] is the correct answer.
Step-by-step explanation:
First of all, let us label the diagram as shown in the attached figure.
OR = 20'
QS = 16'
[tex]\triangle OPQ[/tex] is the right angled triangle that we get.
[tex]\angle P=90^\circ[/tex]
[tex]\angle O =70^\circ[/tex]
From the given figure symmetry, we can clearly see that:
Side OP = OR - QS
OP = 20 - 16 = 4'
Now, let us use trigonometric identity of cosine in [tex]\triangle OPQ[/tex].
[tex]cos\theta = \dfrac{Base}{Hypotenuse}[/tex]
[tex]cosO= \dfrac{OP}{OQ}\\\Rightarrow cos70^\circ= \dfrac{4}{OQ}[/tex]
Let the distance of ramp, OQ = x'
So, the above ratio becomes:
[tex]cos70^\circ= \dfrac{4}{x}[/tex]
So, the correct answer is:
[tex]b.\ cos 70 =\dfrac{4}x[/tex]