Answer:
The distance between the walls is 70 m.
Step-by-step explanation:
Given: A source of laser light is at point A on the ground between two parallel walls BE and CD . The walls are perpendicular to the ground that is
BE ⊥ ED and CD ⊥ ED
AB is a ray of light which strikes the wall on the left at point B which is 30 meters above the ground. that is BE = 30 m
AC is a ray of light which strikes the wall on the right at point C. The length of AC = 80 meters.
The ray AB makes an angle of 45 degrees with the ground that is m∠BAE = 45°
The ray AC makes an angle of 60 degrees with the ground that is m∠CAD = 60°
As shown is figure attached below.
WE have to find the distance between the walls that is Length of ED
Length of ED = EA + AD
Consider the Δ AEB,
Using trigonometric ratio,
![\tan\theta=\frac{perpendicular}{base}](https://tex.z-dn.net/?f=%5Ctan%5Ctheta%3D%5Cfrac%7Bperpendicular%7D%7Bbase%7D)
Here
, perpendicular = 30 m and base we can find.
thus,
![\tan 45^{\circ}=\frac{30}{EA}](https://tex.z-dn.net/?f=%5Ctan%2045%5E%7B%5Ccirc%7D%3D%5Cfrac%7B30%7D%7BEA%7D)
We know ![\tan 45^{\circ}=1](https://tex.z-dn.net/?f=%5Ctan%2045%5E%7B%5Ccirc%7D%3D1)
thus, EA = 30 m
Consider the Δ AEB,
Using trigonometric ratio,
![\cos\theta=\frac{base}{hypotenuse}](https://tex.z-dn.net/?f=%5Ccos%5Ctheta%3D%5Cfrac%7Bbase%7D%7Bhypotenuse%7D)
Here
, hypotenuse = 80 m and base we can find.
thus, ![\cos 60^{\circ}=\frac{base}{80}](https://tex.z-dn.net/?f=%5Ccos%2060%5E%7B%5Ccirc%7D%3D%5Cfrac%7Bbase%7D%7B80%7D)
We know, ![\cos 60^{\circ}=\frac{1}{2}](https://tex.z-dn.net/?f=%5Ccos%2060%5E%7B%5Ccirc%7D%3D%5Cfrac%7B1%7D%7B2%7D)
thus, Base = 40 m
AD = 40 m
Thus, the distance between the walls that is the length of ED = 30 + 40 = 70 m