Question

A source of laser light sends rays AB and AC toward two opposite walls of a hall. The light rays strike the walls at points B and C, as shown below:
A source of laser light is at point A on the ground between two parallel walls. The walls are perpendicular to the ground. AB is a ray of light which strikes the wall on the left at point B which is 30 meters above the ground. AC is a ray of light which strikes the wall on the right at point C. The length of AC is 80 meters. The ray AB makes an angle of 45 degrees with the ground. The ray AC makes an angle of 60 degrees with the ground.

What is the distance between the walls?

Answer 1

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,

$\tan\theta=\frac{perpendicular}{base}$

Here $\theta=45^{\circ}$ , perpendicular = 30 m  and base we can find.

thus,

$\tan 45^{\circ}=\frac{30}{EA}$

We know $\tan 45^{\circ}=1$

thus, EA =  30 m

Consider the Δ AEB,

Using trigonometric ratio,

$\cos\theta=\frac{base}{hypotenuse}$

Here $\theta=60^{\circ}$ , hypotenuse = 80 m  and base we can find.

thus, $\cos 60^{\circ}=\frac{base}{80}$

We know, $\cos 60^{\circ}=\frac{1}{2}$

thus, Base = 40 m

AD = 40 m

Thus, the distance between the walls that is the length of ED = 30 + 40 = 70 m